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

Jan 23, 2018

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

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

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

#### Explanation:

We are given $f \left(x\right) = {\left({e}^{\tan} \left(\frac{1}{x}\right)\right)}^{\frac{1}{2}}$

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

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

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

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

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

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

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

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