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

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Answer 1 · 2018-04-07T22:24:31

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

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

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

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