How do you differentiate f(x) = sqrt(arctan(e^(x-1)) f(x)=arctan(ex1) using the chain rule?

1 Answer
Apr 7, 2018

f'(x)=e^(x-1)/((1+(e^(x-1))^2)2sqrt(arctan(e^(x-1))))

Explanation:

This is a pretty big composition -- the square root, inverse tangent, and exponential will all have to be differentiated.

Differentiate the square root:

f'(x)=1/(2sqrt(arctan(e^(x-1))))*d/dxarctan(e^(x-1))

Recalling that d/dxarctanx=1/(1+x^2),

f'(x)=1/(2sqrt(arctan(e^(x-1))))*1/(1+(e^(x-1))^2)*d/dxe^(x-1)

d/dxe^(x-1)=e^(x-1)*d/dx(x-1)=e^(x-1)

So, we have

f'(x)=e^(x-1)/((1+(e^(x-1))^2)2sqrt(arctan(e^(x-1))))