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

1 Answer
Apr 7, 2018

Answer:

#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))))#