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

Jan 30, 2018

$f ' \left(x\right) = - \frac{{e}^{x} \tan \left({e}^{x}\right)}{2 {\left(\sec \left({e}^{x}\right)\right)}^{\frac{1}{2}}}$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{{e}^{x} \tan \left({e}^{x}\right) \sec {\left({e}^{x}\right)}^{- \frac{1}{2}}}{2}$

#### Explanation:

Here we have $f \left(x\right) = {\left(\sec \left({e}^{x}\right)\right)}^{- \frac{1}{2}}$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[{\left(\sec \left({e}^{x}\right)\right)}^{- \frac{1}{2}}\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\left(\sec \left({e}^{x}\right)\right)}^{- \frac{1}{2} - 1} \cdot - \frac{1}{2} \cdot \frac{d}{\mathrm{dx}} \left[\sec \left({e}^{x}\right)\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\left(\sec \left({e}^{x}\right)\right)}^{- \frac{3}{2}} \cdot - \frac{1}{2} \cdot \frac{d}{\mathrm{dx}} \left[\sec \left({e}^{x}\right)\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\left(\sec \left({e}^{x}\right)\right)}^{- \frac{3}{2}} \cdot - \frac{1}{2} \cdot \sec \left({e}^{x}\right) \tan \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dc}} \left[{e}^{x}\right]$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{{e}^{x} \sec \left({e}^{x}\right) \tan \left({e}^{x}\right)}{2 {\left(\sec \left({e}^{x}\right)\right)}^{\frac{3}{2}}}$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{{e}^{x} \tan \left({e}^{x}\right)}{2 {\left(\sec \left({e}^{x}\right)\right)}^{\frac{1}{2}}}$

$\textcolor{w h i t e}{f ' \left(x\right)} = - \frac{{e}^{x} \tan \left({e}^{x}\right) \sec {\left({e}^{x}\right)}^{- \frac{1}{2}}}{2}$