Find the derivative of the function. Simplify if possible.
and......
y = 17 arctan(sqrt x)
y = arcsin(4x + 2)
y = arccos(e8x)
h(t) = 5arccot(t) + 5arccot(1/t)
Solution
To find the derivative of ???
1)y = 17 arctan sqrtx
tan (y/17) = sqrtx
Differentiating with respect to x, weget:
sec^2 (y/17) * (1/17) dy/dx = (1/2sqrtx)
dy/ dx = (17/2sqrtx){1+(tan(y/17))^2}
dy/dx = (17/2sqrtx) {1+x}
2)y = arc sin(4x+2)
Therefore siny = 4x+2
Differentiate both sides:
cosy*dy/dx = d/dx(4x+2)
cosy * dy/dx = 4
dy/dx = 4/cosy = 4/sin^2y = 4/(4x+2)^2
dy/dx = 4/(4x+2)^2.
3)y=arccos(e^8x).
Therefore cosy = e^(8x).
Differentiating both sodes, we get:
-siny*dy/dx= d/dx{e^(8x)}.
-siny*dy/dx = e^(8x)* d/dx(8x).
-siny*dy/dx = 8e^(8x).
dy/dx = 8e^(8x)* (1/-siny)
dy/dx = -8e^(8x)/ (1-(cosy)^2)
dy/dx = -8e^(8x)/{1- e^(16x)}, as (cosy)^2 = (e^(8x))^2..
