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

Jun 10, 2018

$f ' \left(x\right) = \frac{- {e}^{x - 1}}{2 \left({\left({e}^{x - 1}\right)}^{2} + 1\right) \sqrt{{\left(\arctan \left({e}^{x - 1}\right)\right)}^{3}}}$

#### Explanation:

Let $f \left(x\right) = {\left[\arctan \left({e}^{x - 1}\right)\right]}^{- \frac{1}{2}}$

Use the chain rule, which says that $\textcolor{b l u e}{\frac{d}{\mathrm{dx}} \left(f \left(g \left(x\right)\right)\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)}$:

$f ' \left(x\right) = \left(- \frac{1}{2}\right) {\left(\arctan \left({e}^{x - 1}\right)\right)}^{- \frac{3}{2}} \cdot \left(\frac{1}{{\left({e}^{x - 1}\right)}^{2} + 1}\right) \cdot {e}^{x - 1}$

Now we simplify to get:
$f ' \left(x\right) = \frac{- {e}^{x - 1}}{2 \left({\left({e}^{x - 1}\right)}^{2} + 1\right) \sqrt{{\left(\arctan \left({e}^{x - 1}\right)\right)}^{3}}}$