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Sec on 2.4
    The Product and Quo ent Rules
             V63.0121.011: Calculus I
           Professor Ma hew Leingang
                  New York University


               February 23, 2011


.
Announcements

   Quiz 2 next week on
   §§1.5, 1.6, 2.1, 2.2
   Midterm March 7 on all
   sec ons in class (covers
   all sec ons up to 2.5)
Help!
 Free resources:
      Math Tutoring Center
      (CIWW 524)
      College Learning Center
      (schedule on Blackboard)
      TAs’ office hours
      my office hours
      each other!
Objectives
   Understand and be able
   to use the Product Rule
   for the deriva ve of the
   product of two func ons.
   Understand and be able
   to use the Quo ent Rule
   for the deriva ve of the
   quo ent of two
   func ons.
Outline
 Deriva ve of a Product
    Deriva on
    Examples
 The Quo ent Rule
    Deriva on
    Examples
 More deriva ves of trigonometric func ons
   Deriva ve of Tangent and Cotangent
   Deriva ve of Secant and Cosecant
 More on the Power Rule
   Power Rule for Nega ve Integers
Recollection and extension

 We have shown that if u and v are func ons, that

                         (u + v)′ = u′ + v′
                         (u − v)′ = u′ − v′

 What about uv?
Is the derivative of a product the
product of the derivatives?

              (uv)′ = u′ v′ ?
                    .
Is the derivative of a product the
product of the derivatives?

                              (uv)′ = u′ v′ !
                                    .




 Try this with u = x and v = x2 .
Is the derivative of a product the
product of the derivatives?

                          (uv)′ = u′ v′ !
                                .




 Try this with u = x and v = x2 .
      Then uv = x3 =⇒ (uv)′ = 3x2 .
Is the derivative of a product the
product of the derivatives?

                          (uv)′ = u′ v′ !
                                .




 Try this with u = x and v = x2 .
      Then uv = x3 =⇒ (uv)′ = 3x2 .
      But u′ v′ = 1 · 2x = 2x.
Is the derivative of a product the
product of the derivatives?

                          (uv)′ = u′ v′ !
                                .




 Try this with u = x and v = x2 .
      Then uv = x3 =⇒ (uv)′ = 3x2 .
      But u′ v′ = 1 · 2x = 2x.
 So we have to be more careful.
Mmm...burgers
 Say you work in a fast-food joint. You want to make more money.
 What are your choices?




               .
Mmm...burgers
 Say you work in a fast-food joint. You want to make more money.
 What are your choices?
    Work longer hours.




               .
Mmm...burgers
 Say you work in a fast-food joint. You want to make more money.
 What are your choices?
    Work longer hours.
    Get a raise.




               .
Mmm...burgers
 Say you work in a fast-food joint. You want to make more money.
 What are your choices?
      Work longer hours.
      Get a raise.
 Say you get a 25 cent raise in
 your hourly wages and work 5
 hours more per week. How
 much extra money do you
 make?
                .
Mmm...burgers
 Say you work in a fast-food joint. You want to make more money.
 What are your choices?
      Work longer hours.
      Get a raise.
 Say you get a 25 cent raise in
 your hourly wages and work 5
 hours more per week. How
 much extra money do you
 make?
              .
   ∆I = 5 × $0.25 = $1.25?
Mmm...burgers
 Say you work in a fast-food joint. You want to make more money.
 What are your choices?
      Work longer hours.
      Get a raise.
 Say you get a 25 cent raise in
 your hourly wages and work 5
 hours more per week. How
 much extra money do you
 make?
              .
   ∆I = 5 × $0.25 = $1.25?
Money money money money
 The answer depends on how much you work already and your
 current wage. Suppose you work h hours and are paid w. You get a
  me increase of ∆h and a wage increase of ∆w. Income is wages
  mes hours, so

          ∆I = (w + ∆w)(h + ∆h) − wh
             FOIL
             = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh
             = w · ∆h + ∆w · h + ∆w · ∆h
A geometric argument
 Draw a box:

               ∆h       w ∆h   ∆w ∆h


                h       wh     ∆w h

                    .
                         w      ∆w
A geometric argument
 Draw a box:

               ∆h       w ∆h     ∆w ∆h


                h       wh        ∆w h

                    .
                         w        ∆w

               ∆I = w ∆h + h ∆w + ∆w ∆h
Cash flow
 Supose wages and hours are changing con nuously over me. Over
 a me interval ∆t, what is the average rate of change of income?
                  ∆I   w ∆h + h ∆w + ∆w ∆h
                     =
                  ∆t            ∆t
                        ∆h     ∆w       ∆h
                     =w     +h     + ∆w
                         ∆t     ∆t      ∆t
Cash flow
 Supose wages and hours are changing con nuously over me. Over
 a me interval ∆t, what is the average rate of change of income?
                   ∆I   w ∆h + h ∆w + ∆w ∆h
                      =
                   ∆t            ∆t
                         ∆h     ∆w       ∆h
                      =w     +h     + ∆w
                          ∆t     ∆t      ∆t
 What is the instantaneous rate of change of income?
                 dI       ∆I   dh  dw
                    = lim    =w +h    +0
                 dt ∆t→0 ∆t    dt  dt
Eurekamen!
 We have discovered
 Theorem (The Product Rule)
 Let u and v be differen able at x. Then

                      (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x)

 in Leibniz nota on
                          d         du      dv
                             (uv) =    ·v+u
                          dx        dx      dx
Sanity Check
 Example
 Apply the product rule to u = x and v = x2 .
Sanity Check
 Example
 Apply the product rule to u = x and v = x2 .

 Solu on

      (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2

 This is what we get the “normal” way.
Which is better?

 Example
 Find this deriva ve two ways: first by direct mul plica on and then
 by the product rule:
                      d [                       ]
                          (3 − x2 )(x3 − x + 1)
                      dx
Which is better?
 Example

           d [                       ]
               (3 − x2 )(x3 − x + 1)
           dx
Which is better?
 Example

                           d [                       ]
                               (3 − x2 )(x3 − x + 1)
                           dx

 Solu on
 by direct mul plica on:
            d [                     ] FOIL d [ 5                    ]
               (3 − x2 )(x3 − x + 1) =        −x + 4x3 − x2 − 3x + 3
            dx                            dx
Which is better?
 Example

                           d [                       ]
                               (3 − x2 )(x3 − x + 1)
                           dx

 Solu on
 by direct mul plica on:
            d [                     ] FOIL d [ 5                    ]
               (3 − x2 )(x3 − x + 1) =        −x + 4x3 − x2 − 3x + 3
            dx                            dx
                                       = −5x4 + 12x2 − 2x − 3
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
             = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
             = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
             = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
             = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
             = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
Which is better?
 Example

                         d [                       ]
                             (3 − x2 )(x3 − x + 1)
                         dx

 Solu on
 by the product rule:
                 (            )                       (               )
          dy       d                                    d 3
             =        (3 − x ) (x − x + 1) + (3 − x )
                            2    3                 2
                                                           (x − x + 1)
          dx       dx                                   dx
             = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
             = −5x4 + 12x2 − 2x − 3
One more
 Example
     d
 Find x sin x.
     dx
One more
 Example
     d
 Find x sin x.
     dx
 Solu on

                                (       )          (         )
                 d                  d                d
                    x sin x =          x sin x + x      sin x
                 dx                 dx               dx
One more
 Example
     d
 Find x sin x.
     dx
 Solu on

                              (      )           (          )
                 d               d                  d
                    x sin x =       x sin x + x        sin x
                 dx             dx                  dx
                            = 1 · sin x + x · cos x
One more
 Example
     d
 Find x sin x.
     dx
 Solu on

                              (      )           (          )
                 d               d                  d
                    x sin x =       x sin x + x        sin x
                 dx             dx                  dx
                            = 1 · sin x + x · cos x
                            = sin x + x cos x
Mnemonic
 Let u = “hi” and v = “ho”. Then

           (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
Musical interlude

   jazz bandleader and
   singer
   hit song “Minnie the
   Moocher” featuring “hi
   de ho” chorus
   played Cur s in The Blues
   Brothers
                               Cab Calloway
                                1907–1994
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.

 Solu on

                  (uvw)′                .
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.

 Solu on

                  (uvw)′ = ((uv)w)′ .
Iterating the Product Rule
 Example
                              Apply the product
 Use the product rule to find the deriva ve of a three-fold product uvw.
                              rule to uv and w
 Solu on

                  (uvw)′ = ((uv)w)′ .
Iterating the Product Rule
 Example
                              Apply the product
 Use the product rule to find the deriva ve of a three-fold product uvw.
                              rule to uv and w
 Solu on

                  (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.product
                                                            Apply the
                                                           rule to u and v
 Solu on

                  (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.product
                                                            Apply the
                                                           rule to u and v
 Solu on

                  (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
                         = (u′ v + uv′ )w + (uv)w′
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.

 Solu on

                  (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
                         = (u′ v + uv′ )w + (uv)w′
                         = u′ vw + uv′ w + uvw′
Iterating the Product Rule
 Example
 Use the product rule to find the deriva ve of a three-fold product uvw.

 Solu on

                  (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
                         = (u′ v + uv′ )w + (uv)w′
                         = u′ vw + uv′ w + uvw′


 So we write down the product three mes, taking the deriva ve of each factor
 once.
Outline
 Deriva ve of a Product
    Deriva on
    Examples
 The Quo ent Rule
    Deriva on
    Examples
 More deriva ves of trigonometric func ons
   Deriva ve of Tangent and Cotangent
   Deriva ve of Secant and Cosecant
 More on the Power Rule
   Power Rule for Nega ve Integers
The Quotient Rule
 What about the deriva ve of a quo ent?
The Quotient Rule
 What about the deriva ve of a quo ent?
                                                 u
 Let u and v be differen able func ons and let Q = . Then
                                                 v
                               u = Qv
The Quotient Rule
 What about the deriva ve of a quo ent?
                                                 u
 Let u and v be differen able func ons and let Q = . Then
                                                 v
                               u = Qv
 If Q is differen able, we have
                            u′ = (Qv)′ = Q′ v + Qv′
The Quotient Rule
 What about the deriva ve of a quo ent?
                                                 u
 Let u and v be differen able func ons and let Q = . Then
                                                 v
                               u = Qv
 If Q is differen able, we have
                          u′ = (Qv)′ = Q′ v + Qv′
                           ′   u′ − Qv′     u′ u v′
                       =⇒ Q =           = − ·
                                   v        v   v v
The Quotient Rule
 What about the deriva ve of a quo ent?
                                                 u
 Let u and v be differen able func ons and let Q = . Then
                                                 v
                               u = Qv
 If Q is differen able, we have
                          u′ = (Qv)′ = Q′ v + Qv′
                           ′   u′ − Qv′     u′ u v′
                    =⇒ Q =              = − ·
                                   v        v   v v
                      ( u )′ u′ v − uv′
              =⇒ Q′ =        =
                        v          v2
The Quotient Rule
 What about the deriva ve of a quo ent?
                                                 u
 Let u and v be differen able func ons and let Q = . Then
                                                 v
                               u = Qv
 If Q is differen able, we have
                             u′ = (Qv)′ = Q′ v + Qv′
                              ′   u′ − Qv′     u′ u v′
                        =⇒ Q =             = − ·
                                      v        v   v v
                         ( u )′ u′ v − uv′
                 =⇒ Q′ =        =
                           v          v2
 This is called the Quo ent Rule.
The Quotient Rule
 We have discovered
 Theorem (The Quo ent Rule)
                                                      u
 Let u and v be differen able at x, and v(x) ̸= 0. Then is
                                                      v
 differen able at x, and
                  ( u )′           u′ (x)v(x) − u(x)v′ (x)
                           (x) =
                    v                       v(x)2
Verifying Example
 Example
                                          (        )
                                     d        x2                             d
 Verify the quo ent rule by compu ng                   and comparing it to      (x).
                                     dx       x                              dx
Verifying Example
 Example
                                          (        )
                                     d        x2                             d
 Verify the quo ent rule by compu ng                   and comparing it to      (x).
                                     dx       x                              dx

 Solu on
                 ( 2)       ( )
            d     x     x dx x2 − x2 dx (x) x · 2x − x2 · 1
                          d           d
                      =                    =
            dx     x            x2                 x2
                        x2         d
                      = 2 =1=         (x)
                        x          dx
Mnemonic
 Let u = “hi” and v = “lo”. Then
       ( u )′ vu′ − uv′
             =           = “lo dee hi minus hi dee lo over lo lo”
         v        v2
Examples

 Example
     d 2x + 5
  1.
     dx 3x − 2
     d sin x
  2.
     dx x2
     d      1
  3.     2+t+2
     dt t
Solution to first example
 Solu on


      d 2x + 5
      dx 3x − 2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                        d                      d
                =
      dx 3x − 2                 (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                        d                      d
                =
      dx 3x − 2                 (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                        d                      d
                =
      dx 3x − 2                 (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                        d                      d
                =
      dx 3x − 2                 (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                        d                      d
                =
      dx 3x − 2                 (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                        d                      d
                =
      dx 3x − 2                 (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                   d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                   d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                   d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                   d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                   d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                   d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
                  (6x − 4) − (6x + 15)
                =
                       (3x − 2)2
Solution to first example
 Solu on


      d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2)
                           d                    d
                =
      dx 3x − 2                  (3x − 2)2
                  (3x − 2)(2) − (2x + 5)(3)
                =
                          (3x − 2)2
                  (6x − 4) − (6x + 15)         19
                =                      =−
                       (3x − 2)2            (3x − 2)2
Examples

 Example         Answers
     d 2x + 5               19
  1.              1. −
     dx 3x − 2           (3x − 2)2
     d sin x
  2.
     dx x2
     d      1
  3.     2+t+2
     dt t
Solution to second example
 Solu on


           d sin x
                   =
           dx x2
Solution to second example
 Solu on


           d sin x x2
                  =
           dx x2
Solution to second example
 Solu on


                      d
           d sin x x2 dx sin x
                  =
           dx x2
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x
                      d
                  =
           dx x2
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                      d                d
                  =
           dx x2
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                      d                d
                  =
           dx x2             (x2 )2
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                      d                d
                  =
           dx x2             (x2 )2

                   =
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                       d               d
                  =
           dx x2             (x2 )2
                    x2
                  =
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                       d               d
                  =
           dx x2             (x2 )2
                    x2 cos x
                  =
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                       d               d
                  =
           dx x2             (x2 )2
                    x2 cos x − 2x
                  =
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                       d               d
                  =
           dx x2             (x2 )2
                    x2 cos x − 2x sin x
                  =
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                       d               d
                  =
           dx x2             (x2 )2
                    x2 cos x − 2x sin x
                  =
                             x4
Solution to second example
 Solu on


           d sin x x2 dx sin x − sin x dx x2
                       d               d
                  =
           dx x2             (x2 )2
                    x2 cos x − 2x sin x
                  =
                             x4
                    x cos x − 2 sin x
                  =
                            x3
Another way to do it
 Find the deriva ve with the product rule instead.
 Solu on
                d sin x    d (             )
                     2
                        =      sin x · x−2
                dx x      dx
                          (          )               (        )
                             d             −2            d −2
                        =       sin x · x + sin x ·         x
                            dx                           dx
                        = cos x · x−2 + sin x · (−2x−3 )
                        = x−3 (x cos x − 2 sin x)


 No ce the technique of factoring out the largest nega ve power,
 leaving posi ve powers.
Examples

 Example         Answers
     d 2x + 5              19
  1.              1. −
     dx 3x − 2         (3x − 2)2
     d sin x         x cos x − 2 sin x
  2.              2.
     dx x2                   x3
     d      1
  3.     2+t+2
     dt t
Solution to third example
 Solu on


           d      1
           dt t2 + t + 2
Solution to third example
 Solu on


           d      1        (t2 + t + 2)(0) − (1)(2t + 1)
                         =
           dt t2 + t + 2            (t2 + t + 2)2
Solution to third example
 Solu on


           d      1        (t2 + t + 2)(0) − (1)(2t + 1)
                         =
           dt t2 + t + 2            (t2 + t + 2)2
                                2t + 1
                         =− 2
                             (t + t + 2)2
A nice little takeaway
 Fact
                                                1
 Let v be differen able at x, and v(x) ̸= 0. Then is differen able at
                                                v
 0, and                      ( )′
                               1         v′
                                   =− 2
                               v         v

 Proof.
               ( )
          d     1    v·   d
                          dx (1)−1·   d
                                      dx v     v · 0 − 1 · v′   v′
                   =                         =                =− 2
          dx    v              v2                    v2         v
Examples

 Example         Answers
     d 2x + 5              19
  1.              1. −
     dx 3x − 2         (3x − 2)2
     d sin x         x cos x − 2 sin x
  2.              2.
     dx x2                   x3
     d      1              2t + 1
  3.              3. − 2
     dt t2+t+2         (t + t + 2)2
Outline
 Deriva ve of a Product
    Deriva on
    Examples
 The Quo ent Rule
    Deriva on
    Examples
 More deriva ves of trigonometric func ons
   Deriva ve of Tangent and Cotangent
   Deriva ve of Secant and Cosecant
 More on the Power Rule
   Power Rule for Nega ve Integers
Derivative of Tangent
 Example
      d
 Find    tan x
      dx
Derivative of Tangent
 Example
      d
 Find    tan x
      dx

 Solu on

                       (           )
       d          d        sin x
          tan x =
       dx         dx       cos x
Derivative of Tangent
 Example
      d
 Find    tan x
      dx

 Solu on

                       (           )
       d          d        sin x           cos x · cos x − sin x · (− sin x)
          tan x =                      =
       dx         dx       cos x                        cos2 x
Derivative of Tangent
 Example
      d
 Find    tan x
      dx

 Solu on

                     (       )
       d           d sin x        cos x · cos x − sin x · (− sin x)
          tan x =               =
       dx         dx cos x                     cos2 x
                  cos2 x + sin2 x
                =
                      cos2 x
Derivative of Tangent
 Example
      d
 Find    tan x
      dx

 Solu on

                     (       )
       d           d sin x        cos x · cos x − sin x · (− sin x)
          tan x =               =
       dx         dx cos x                     cos2 x
                  cos2 x + sin2 x     1
                =                 =
                      cos2 x        cos2 x
Derivative of Tangent
 Example
      d
 Find    tan x
      dx

 Solu on

                     (       )
       d           d sin x        cos x · cos x − sin x · (− sin x)
          tan x =               =
       dx         dx cos x                     cos2 x
                  cos2 x + sin2 x     1
                =         2x
                                  =     2x
                                            = sec2 x
                      cos           cos
Derivative of Cotangent
 Example
      d
 Find    cot x
      dx
Derivative of Cotangent
 Example
      d
 Find    cot x
      dx
 Answer

                 d             1
                    cot x = − 2 = − csc2 x
                 dx          sin x
Derivative of Cotangent
 Example
      d
 Find    cot x
      dx

 Solu on


       d          d ( cos x )   sin x · (− sin x) − cos x · cos x
          cot x =             =
       dx         dx sin x                    sin2 x
Derivative of Cotangent
 Example
      d
 Find    cot x
      dx

 Solu on


       d          d ( cos x )    sin x · (− sin x) − cos x · cos x
          cot x =             =
       dx         dx sin x                     sin2 x
                  − sin2 x − cos2 x
                =
                       sin2 x
Derivative of Cotangent
 Example
      d
 Find    cot x
      dx

 Solu on


       d          d ( cos x )    sin x · (− sin x) − cos x · cos x
          cot x =             =
       dx         dx sin x                     sin2 x
                  − sin2 x − cos2 x         1
                =                   =− 2
                       sin2 x             sin x
Derivative of Cotangent
 Example
      d
 Find    cot x
      dx

 Solu on


       d          d ( cos x )    sin x · (− sin x) − cos x · cos x
          cot x =             =
       dx         dx sin x                     sin2 x
                  − sin2 x − cos2 x         1
                =                   = − 2 = − csc2 x
                       sin2 x             sin x
Derivative of Secant
 Example
      d
 Find    sec x
      dx
Derivative of Secant
 Example
      d
 Find    sec x
      dx

 Solu on

                           (           )
           d          d          1
              sec x =
           dx         dx       cos x
Derivative of Secant
 Example
      d
 Find    sec x
      dx

 Solu on

                           (           )
           d          d          1             cos x · 0 − 1 · (− sin x)
              sec x =                      =
           dx         dx       cos x                     cos2 x
Derivative of Secant
 Example
      d
 Find    sec x
      dx

 Solu on

                          (    )
           d           d     1     cos x · 0 − 1 · (− sin x)
              sec x =            =
           dx         dx cos x               cos2 x
                       sin x
                    =
                      cos2 x
Derivative of Secant
 Example
      d
 Find    sec x
      dx

 Solu on

                          (    )
           d           d     1        cos x · 0 − 1 · (− sin x)
              sec x =             =
           dx         dx cos x                  cos2 x
                       sin x    1      sin x
                    =        =      ·
                      cos2 x cos x cos x
Derivative of Secant
 Example
      d
 Find    sec x
      dx

 Solu on

                          (    )
           d           d     1        cos x · 0 − 1 · (− sin x)
              sec x =             =
           dx         dx cos x                  cos2 x
                       sin x    1      sin x
                    =        =      ·         = sec x tan x
                      cos2 x cos x cos x
Derivative of Cosecant
 Example
      d
 Find    csc x
      dx
Derivative of Cosecant
 Example
      d
 Find    csc x
      dx
 Answer

                 d
                    csc x = − csc x cot x
                 dx
Derivative of Cosecant
 Example
      d
 Find    csc x
      dx

 Solu on

                           (           )
           d          d          1             sin x · 0 − 1 · (cos x)
              csc x =                      =
           dx         dx       sin x                    sin2 x
Derivative of Cosecant
 Example
      d
 Find    csc x
      dx

 Solu on

                         (     )
           d          d      1     sin x · 0 − 1 · (cos x)
              csc x =            =
           dx         dx sin x              sin2 x
                        cos x
                    =− 2
                        sin x
Derivative of Cosecant
 Example
      d
 Find    csc x
      dx

 Solu on

                         (     )
           d          d      1      sin x · 0 − 1 · (cos x)
              csc x =            =
           dx         dx sin x               sin2 x
                        cos x      1     cos x
                    =− 2 =−            ·
                        sin x    sin x sin x
Derivative of Cosecant
 Example
      d
 Find    csc x
      dx

 Solu on

                         (     )
           d          d      1      sin x · 0 − 1 · (cos x)
              csc x =            =
           dx         dx sin x               sin2 x
                        cos x      1     cos x
                    =− 2 =−            ·        = − csc x cot x
                        sin x    sin x sin x
Recap: Derivatives of
trigonometric functions
    y          y′
   sin x      cos x      Func ons come in pairs
                         (sin/cos, tan/cot, sec/csc)
   cos x    − sin x
                         Deriva ves of pairs follow
   tan x      sec2 x     similar pa erns, with
   cot x   − csc2 x      func ons and
                         co-func ons switched
   sec x   sec x tan x   and an extra sign.
   csc x − csc x cot x
Outline
 Deriva ve of a Product
    Deriva on
    Examples
 The Quo ent Rule
    Deriva on
    Examples
 More deriva ves of trigonometric func ons
   Deriva ve of Tangent and Cotangent
   Deriva ve of Secant and Cosecant
 More on the Power Rule
   Power Rule for Nega ve Integers
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.

 Proof.
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.

 Proof.

    d −n   d 1
       x =
    dx     dx xn
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.

 Proof.
                    d n
    d −n   d 1      dx x
       x =       =− n 2
    dx     dx xn   (x )
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.

 Proof.
                     d n
    d −n   d 1       dx x   nxn−1
       x =       = − n 2 = − 2n
    dx     dx xn    (x )     x
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.

 Proof.
                     d n
    d −n   d 1       dx x   nxn−1
       x =     n
                 = − n 2 = − 2n = −nxn−1−2n
    dx     dx x     (x )     x
Power Rule for Negative Integers
 We will use the quo ent rule to prove
 Theorem

                           d −n
                              x = (−n)x−n−1
                           dx
 for posi ve integers n.

 Proof.
                     d n
    d −n   d 1       dx x   nxn−1
       x =     n
                 = − n 2 = − 2n = −nxn−1−2n = −nx−n−1
    dx     dx x     (x )     x
Summary
  The Product Rule: (uv)′ = u′ v + uv′
                     ( u )′ vu′ − uv′
  The Quo ent Rule:        =
                       v          v2
  Deriva ves of tangent/cotangent, secant/cosecant
         d                           d
           tan x = sec2 x              sec x = sec x tan x
        dx                          dx
         d                           d
           cot x = − csc2 x            csc x = − csc x cot x
        dx                          dx
  The Power Rule is true for all whole number powers, including
  nega ve powers:
                            d n
                              x = nxn−1
                           dx

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Lesson 9: The Product and Quotient Rules (slides)

  • 1. Sec on 2.4 The Product and Quo ent Rules V63.0121.011: Calculus I Professor Ma hew Leingang New York University February 23, 2011 .
  • 2. Announcements Quiz 2 next week on §§1.5, 1.6, 2.1, 2.2 Midterm March 7 on all sec ons in class (covers all sec ons up to 2.5)
  • 3. Help! Free resources: Math Tutoring Center (CIWW 524) College Learning Center (schedule on Blackboard) TAs’ office hours my office hours each other!
  • 4. Objectives Understand and be able to use the Product Rule for the deriva ve of the product of two func ons. Understand and be able to use the Quo ent Rule for the deriva ve of the quo ent of two func ons.
  • 5. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
  • 6. Recollection and extension We have shown that if u and v are func ons, that (u + v)′ = u′ + v′ (u − v)′ = u′ − v′ What about uv?
  • 7. Is the derivative of a product the product of the derivatives? (uv)′ = u′ v′ ? .
  • 8. Is the derivative of a product the product of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 .
  • 9. Is the derivative of a product the product of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 .
  • 10. Is the derivative of a product the product of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x.
  • 11. Is the derivative of a product the product of the derivatives? (uv)′ = u′ v′ ! . Try this with u = x and v = x2 . Then uv = x3 =⇒ (uv)′ = 3x2 . But u′ v′ = 1 · 2x = 2x. So we have to be more careful.
  • 12. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? .
  • 13. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. .
  • 14. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. .
  • 15. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? .
  • 16. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . ∆I = 5 × $0.25 = $1.25?
  • 17. Mmm...burgers Say you work in a fast-food joint. You want to make more money. What are your choices? Work longer hours. Get a raise. Say you get a 25 cent raise in your hourly wages and work 5 hours more per week. How much extra money do you make? . ∆I = 5 × $0.25 = $1.25?
  • 18. Money money money money The answer depends on how much you work already and your current wage. Suppose you work h hours and are paid w. You get a me increase of ∆h and a wage increase of ∆w. Income is wages mes hours, so ∆I = (w + ∆w)(h + ∆h) − wh FOIL = w · h + w · ∆h + ∆w · h + ∆w · ∆h − wh = w · ∆h + ∆w · h + ∆w · ∆h
  • 19. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h . w ∆w
  • 20. A geometric argument Draw a box: ∆h w ∆h ∆w ∆h h wh ∆w h . w ∆w ∆I = w ∆h + h ∆w + ∆w ∆h
  • 21. Cash flow Supose wages and hours are changing con nuously over me. Over a me interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t
  • 22. Cash flow Supose wages and hours are changing con nuously over me. Over a me interval ∆t, what is the average rate of change of income? ∆I w ∆h + h ∆w + ∆w ∆h = ∆t ∆t ∆h ∆w ∆h =w +h + ∆w ∆t ∆t ∆t What is the instantaneous rate of change of income? dI ∆I dh dw = lim =w +h +0 dt ∆t→0 ∆t dt dt
  • 23. Eurekamen! We have discovered Theorem (The Product Rule) Let u and v be differen able at x. Then (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) in Leibniz nota on d du dv (uv) = ·v+u dx dx dx
  • 24. Sanity Check Example Apply the product rule to u = x and v = x2 .
  • 25. Sanity Check Example Apply the product rule to u = x and v = x2 . Solu on (uv)′ (x) = u(x)v′ (x) + u′ (x)v(x) = x · (2x) + 1 · x2 = 3x2 This is what we get the “normal” way.
  • 26. Which is better? Example Find this deriva ve two ways: first by direct mul plica on and then by the product rule: d [ ] (3 − x2 )(x3 − x + 1) dx
  • 27. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx
  • 28. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by direct mul plica on: d [ ] FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx
  • 29. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by direct mul plica on: d [ ] FOIL d [ 5 ] (3 − x2 )(x3 − x + 1) = −x + 4x3 − x2 − 3x + 3 dx dx = −5x4 + 12x2 − 2x − 3
  • 30. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx
  • 31. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
  • 32. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
  • 33. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
  • 34. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
  • 35. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1)
  • 36. Which is better? Example d [ ] (3 − x2 )(x3 − x + 1) dx Solu on by the product rule: ( ) ( ) dy d d 3 = (3 − x ) (x − x + 1) + (3 − x ) 2 3 2 (x − x + 1) dx dx dx = (−2x)(x3 − x + 1) + (3 − x2 )(3x2 − 1) = −5x4 + 12x2 − 2x − 3
  • 37. One more Example d Find x sin x. dx
  • 38. One more Example d Find x sin x. dx Solu on ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx
  • 39. One more Example d Find x sin x. dx Solu on ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x
  • 40. One more Example d Find x sin x. dx Solu on ( ) ( ) d d d x sin x = x sin x + x sin x dx dx dx = 1 · sin x + x · cos x = sin x + x cos x
  • 41. Mnemonic Let u = “hi” and v = “ho”. Then (uv)′ = vu′ + uv′ = “ho dee hi plus hi dee ho”
  • 42. Musical interlude jazz bandleader and singer hit song “Minnie the Moocher” featuring “hi de ho” chorus played Cur s in The Blues Brothers Cab Calloway 1907–1994
  • 43. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw.
  • 44. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw. Solu on (uvw)′ .
  • 45. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw. Solu on (uvw)′ = ((uv)w)′ .
  • 46. Iterating the Product Rule Example Apply the product Use the product rule to find the deriva ve of a three-fold product uvw. rule to uv and w Solu on (uvw)′ = ((uv)w)′ .
  • 47. Iterating the Product Rule Example Apply the product Use the product rule to find the deriva ve of a three-fold product uvw. rule to uv and w Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
  • 48. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw.product Apply the rule to u and v Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ .
  • 49. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw.product Apply the rule to u and v Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′
  • 50. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw. Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′
  • 51. Iterating the Product Rule Example Use the product rule to find the deriva ve of a three-fold product uvw. Solu on (uvw)′ = ((uv)w)′ . = (uv)′ w + (uv)w′ . = (u′ v + uv′ )w + (uv)w′ = u′ vw + uv′ w + uvw′ So we write down the product three mes, taking the deriva ve of each factor once.
  • 52. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
  • 53. The Quotient Rule What about the deriva ve of a quo ent?
  • 54. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be differen able func ons and let Q = . Then v u = Qv
  • 55. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be differen able func ons and let Q = . Then v u = Qv If Q is differen able, we have u′ = (Qv)′ = Q′ v + Qv′
  • 56. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be differen able func ons and let Q = . Then v u = Qv If Q is differen able, we have u′ = (Qv)′ = Q′ v + Qv′ ′ u′ − Qv′ u′ u v′ =⇒ Q = = − · v v v v
  • 57. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be differen able func ons and let Q = . Then v u = Qv If Q is differen able, we have u′ = (Qv)′ = Q′ v + Qv′ ′ u′ − Qv′ u′ u v′ =⇒ Q = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2
  • 58. The Quotient Rule What about the deriva ve of a quo ent? u Let u and v be differen able func ons and let Q = . Then v u = Qv If Q is differen able, we have u′ = (Qv)′ = Q′ v + Qv′ ′ u′ − Qv′ u′ u v′ =⇒ Q = = − · v v v v ( u )′ u′ v − uv′ =⇒ Q′ = = v v2 This is called the Quo ent Rule.
  • 59. The Quotient Rule We have discovered Theorem (The Quo ent Rule) u Let u and v be differen able at x, and v(x) ̸= 0. Then is v differen able at x, and ( u )′ u′ (x)v(x) − u(x)v′ (x) (x) = v v(x)2
  • 60. Verifying Example Example ( ) d x2 d Verify the quo ent rule by compu ng and comparing it to (x). dx x dx
  • 61. Verifying Example Example ( ) d x2 d Verify the quo ent rule by compu ng and comparing it to (x). dx x dx Solu on ( 2) ( ) d x x dx x2 − x2 dx (x) x · 2x − x2 · 1 d d = = dx x x2 x2 x2 d = 2 =1= (x) x dx
  • 62. Mnemonic Let u = “hi” and v = “lo”. Then ( u )′ vu′ − uv′ = = “lo dee hi minus hi dee lo over lo lo” v v2
  • 63. Examples Example d 2x + 5 1. dx 3x − 2 d sin x 2. dx x2 d 1 3. 2+t+2 dt t
  • 64. Solution to first example Solu on d 2x + 5 dx 3x − 2
  • 65. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
  • 66. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
  • 67. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
  • 68. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
  • 69. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
  • 70. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2
  • 71. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
  • 72. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
  • 73. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
  • 74. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
  • 75. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2
  • 76. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) = (3x − 2)2
  • 77. Solution to first example Solu on d 2x + 5 (3x − 2) dx (2x + 5) − (2x + 5) dx (3x − 2) d d = dx 3x − 2 (3x − 2)2 (3x − 2)(2) − (2x + 5)(3) = (3x − 2)2 (6x − 4) − (6x + 15) 19 = =− (3x − 2)2 (3x − 2)2
  • 78. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x 2. dx x2 d 1 3. 2+t+2 dt t
  • 79. Solution to second example Solu on d sin x = dx x2
  • 80. Solution to second example Solu on d sin x x2 = dx x2
  • 81. Solution to second example Solu on d d sin x x2 dx sin x = dx x2
  • 82. Solution to second example Solu on d sin x x2 dx sin x − sin x d = dx x2
  • 83. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2
  • 84. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2
  • 85. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 =
  • 86. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 =
  • 87. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x =
  • 88. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x =
  • 89. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x sin x =
  • 90. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x sin x = x4
  • 91. Solution to second example Solu on d sin x x2 dx sin x − sin x dx x2 d d = dx x2 (x2 )2 x2 cos x − 2x sin x = x4 x cos x − 2 sin x = x3
  • 92. Another way to do it Find the deriva ve with the product rule instead. Solu on d sin x d ( ) 2 = sin x · x−2 dx x dx ( ) ( ) d −2 d −2 = sin x · x + sin x · x dx dx = cos x · x−2 + sin x · (−2x−3 ) = x−3 (x cos x − 2 sin x) No ce the technique of factoring out the largest nega ve power, leaving posi ve powers.
  • 93. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x x cos x − 2 sin x 2. 2. dx x2 x3 d 1 3. 2+t+2 dt t
  • 94. Solution to third example Solu on d 1 dt t2 + t + 2
  • 95. Solution to third example Solu on d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2
  • 96. Solution to third example Solu on d 1 (t2 + t + 2)(0) − (1)(2t + 1) = dt t2 + t + 2 (t2 + t + 2)2 2t + 1 =− 2 (t + t + 2)2
  • 97. A nice little takeaway Fact 1 Let v be differen able at x, and v(x) ̸= 0. Then is differen able at v 0, and ( )′ 1 v′ =− 2 v v Proof. ( ) d 1 v· d dx (1)−1· d dx v v · 0 − 1 · v′ v′ = = =− 2 dx v v2 v2 v
  • 98. Examples Example Answers d 2x + 5 19 1. 1. − dx 3x − 2 (3x − 2)2 d sin x x cos x − 2 sin x 2. 2. dx x2 x3 d 1 2t + 1 3. 3. − 2 dt t2+t+2 (t + t + 2)2
  • 99. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
  • 100. Derivative of Tangent Example d Find tan x dx
  • 101. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x tan x = dx dx cos x
  • 102. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x
  • 103. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x = cos2 x
  • 104. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = = cos2 x cos2 x
  • 105. Derivative of Tangent Example d Find tan x dx Solu on ( ) d d sin x cos x · cos x − sin x · (− sin x) tan x = = dx dx cos x cos2 x cos2 x + sin2 x 1 = 2x = 2x = sec2 x cos cos
  • 106. Derivative of Cotangent Example d Find cot x dx
  • 107. Derivative of Cotangent Example d Find cot x dx Answer d 1 cot x = − 2 = − csc2 x dx sin x
  • 108. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x
  • 109. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x = sin2 x
  • 110. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x 1 = =− 2 sin2 x sin x
  • 111. Derivative of Cotangent Example d Find cot x dx Solu on d d ( cos x ) sin x · (− sin x) − cos x · cos x cot x = = dx dx sin x sin2 x − sin2 x − cos2 x 1 = = − 2 = − csc2 x sin2 x sin x
  • 112. Derivative of Secant Example d Find sec x dx
  • 113. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 sec x = dx dx cos x
  • 114. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x
  • 115. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x = cos2 x
  • 116. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · cos2 x cos x cos x
  • 117. Derivative of Secant Example d Find sec x dx Solu on ( ) d d 1 cos x · 0 − 1 · (− sin x) sec x = = dx dx cos x cos2 x sin x 1 sin x = = · = sec x tan x cos2 x cos x cos x
  • 118. Derivative of Cosecant Example d Find csc x dx
  • 119. Derivative of Cosecant Example d Find csc x dx Answer d csc x = − csc x cot x dx
  • 120. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x
  • 121. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x =− 2 sin x
  • 122. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · sin x sin x sin x
  • 123. Derivative of Cosecant Example d Find csc x dx Solu on ( ) d d 1 sin x · 0 − 1 · (cos x) csc x = = dx dx sin x sin2 x cos x 1 cos x =− 2 =− · = − csc x cot x sin x sin x sin x
  • 124. Recap: Derivatives of trigonometric functions y y′ sin x cos x Func ons come in pairs (sin/cos, tan/cot, sec/csc) cos x − sin x Deriva ves of pairs follow tan x sec2 x similar pa erns, with cot x − csc2 x func ons and co-func ons switched sec x sec x tan x and an extra sign. csc x − csc x cot x
  • 125. Outline Deriva ve of a Product Deriva on Examples The Quo ent Rule Deriva on Examples More deriva ves of trigonometric func ons Deriva ve of Tangent and Cotangent Deriva ve of Secant and Cosecant More on the Power Rule Power Rule for Nega ve Integers
  • 126. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n.
  • 127. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof.
  • 128. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d −n d 1 x = dx dx xn
  • 129. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x x = =− n 2 dx dx xn (x )
  • 130. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x nxn−1 x = = − n 2 = − 2n dx dx xn (x ) x
  • 131. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x nxn−1 x = n = − n 2 = − 2n = −nxn−1−2n dx dx x (x ) x
  • 132. Power Rule for Negative Integers We will use the quo ent rule to prove Theorem d −n x = (−n)x−n−1 dx for posi ve integers n. Proof. d n d −n d 1 dx x nxn−1 x = n = − n 2 = − 2n = −nxn−1−2n = −nx−n−1 dx dx x (x ) x
  • 133. Summary The Product Rule: (uv)′ = u′ v + uv′ ( u )′ vu′ − uv′ The Quo ent Rule: = v v2 Deriva ves of tangent/cotangent, secant/cosecant d d tan x = sec2 x sec x = sec x tan x dx dx d d cot x = − csc2 x csc x = − csc x cot x dx dx The Power Rule is true for all whole number powers, including nega ve powers: d n x = nxn−1 dx