calculuc

1. Made by:
Rudra Patel
Harsh Desai
Ravi Patel
Rishabh Patel
Harshil Raymagiya

2. We now have a pretty good list of “shortcuts” to find
derivatives of simple functions.
Of course, many of the functions that we will encounter
are not so simple. What is needed is a way to combine
derivative rules to evaluate more complicated functions.
®

3. Consider a simple composite function:
y = 6x -10
y = 2( 3x -5)
If u = 3x -5
then y = 2u
y = 6x -10 y = 2u u = 3x - 5
= 2 dy
dy 6
dx
du
= du 3
dx
=
6 = 2×3
dy = dy ×
du
dx du dx
®

4. and another:
y = 5u - 2
where u = 3t
then y = 5( 3t ) - 2
u = 3t
dy 15
dt
= dy 5
du
= du 3
dt
=
15 = 5×3
dy = dy ×
du
dt du dt
y = 5( 3t ) - 2
y =15t - 2
y = 5u - 2
®

5. and one more:
y = 9x2 + 6x +1
( ) 2 y = 3x +1
If u = 3x +1
u = 3x +1
dy 18x 6
dx
y = u2
= + dy 2u
du
= du 3
dx
=
dy = dy ×
du
dx du dx
then y = u2
y = 9x2 + 6x +1
dy 2( 3x 1)
du
= +
dy = 6x +
2
du
18x + 6 = ( 6x + 2) ×3
This pattern is called
the chain rule.
®

6. dy dy du
dx du dx
Chain Rule: = ×
f o g y = f ( u) u = g ( x)
If is the composite of and ,
then:
o ¢ = ¢ × ¢
( ) at u g( x) at x f g f g =
example: f ( x) = sin x g ( x) = x2 - 4 Find: ( f o g )¢ at x = 2
f ¢( x) = cos x g¢( x) = 2x g ( 2) = 4 - 4 = 0
f ¢( 0) × g¢( 2)
cos ( 0) ×( 2×2)
1× 4 = 4
®

7. We could also do it this way:
f ( g ( x) ) = sin ( x2 - 4)
y = sin ( x2 - 4)
y = sin u u = x2 - 4
dy = cosu
du =
2x
du
dx
dy = dy ×
du
dx du dx
dy = cosu ×
2x
dx
dy = cos ( x2 - 4) ×
2x
dx
dy = cos ( 22 - 4) × 2 ×
2
dx
dy = cos( 0) ×
4
dx
dy =
4
dx
®

8. Here is a faster way to find the derivative:
y = sin ( x2 - 4)
y ¢ = cos( x2 - 4) ´ d ( x2 -
4)
dx
y¢ = cos ( x2 - 4) ´2x
Differentiate the outside function...
…then the inside function
At x = 2, y¢ = 4
®

9. Another example:
d cos2 ( 3x)
dx
d éë cos ( 3x
) 2 dx
ùû
2 cos ( 3x) d cos ( 3x)
dx
éë ùû ×
derivative of the
outside function
derivative of the
inside function
It looks like we need to
use the chain rule again!
®

10. Another example:
d cos2 ( 3x)
dx
d éë cos ( 3x
) 2 dx
ùû
2 cos ( 3x) d cos ( 3x)
dx
éë ùû ×
2cos ( 3x) sin ( 3x) d ( 3x)
dx
×- ×
-2cos ( 3x) ×sin ( 3x) ×3
-6cos ( 3x) sin ( 3x)
The chain rule can be used
more than once.
(That’s what makes the
“chain” in the “chain rule”!)
®

11. Derivative formulas include the chain rule!
d un = nun - 1 du
d sin u =
cosu du
dx dx
dx dx
d cosu sin u du
dx dx
= - d tan u =
sec2 u du
dx dx
etcetera…
The formulas on the memorization sheet are written with
instead of . Don’t forget to include the term!
u¢
du u¢
dx
®

12. The most common mistake on the chapter 3 test is to
forget to use the chain rule.
Every derivative problem could be thought of as a chain-rule
problem:
d x2
dx
2x d x
dx
= = 2x ×1 = 2x
derivative of
outside function
derivative of
inside function
The derivative of x is one.
®

13. The chain rule enables us to find the slope of
parametrically defined curves:
dy = dy ×
dx
dt dx dt
dy
dt dy
dx =
dx
dt
Divide both sides by
dx
The slope of a pardatmetrized
curve is given by:
dy
dy dt
dx dx
dt
=
®

14. Example: x = 3cos t y = 2sin t
These are the equations for
an ellipse.
dx 3sin t
dt
= - dy 2cos t
dt
dy t
dx t
= 2cos
=
-
3sin
2 cot
3
= - t

15. p