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

Apr 7, 2018

$f ' \left(x\right) = {e}^{x - 1} / \left(\left(1 + {\left({e}^{x - 1}\right)}^{2}\right) 2 \sqrt{\arctan \left({e}^{x - 1}\right)}\right)$

Explanation:

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

Differentiate the square root:

$f ' \left(x\right) = \frac{1}{2 \sqrt{\arctan \left({e}^{x - 1}\right)}} \cdot \frac{d}{\mathrm{dx}} \arctan \left({e}^{x - 1}\right)$

Recalling that $\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}} ,$

$f ' \left(x\right) = \frac{1}{2 \sqrt{\arctan \left({e}^{x - 1}\right)}} \cdot \frac{1}{1 + {\left({e}^{x - 1}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} {e}^{x - 1}$

$\frac{d}{\mathrm{dx}} {e}^{x - 1} = {e}^{x - 1} \cdot \frac{d}{\mathrm{dx}} \left(x - 1\right) = {e}^{x - 1}$

So, we have

$f ' \left(x\right) = {e}^{x - 1} / \left(\left(1 + {\left({e}^{x - 1}\right)}^{2}\right) 2 \sqrt{\arctan \left({e}^{x - 1}\right)}\right)$