# How do you differentiate f(x)=csc(sqrt(e^x))  using the chain rule?

Jan 5, 2016

$f ' \left(x\right) = \frac{- \sqrt{{e}^{x}} \csc \left(\sqrt{{e}^{x}}\right) \cot \left(\sqrt{{e}^{x}}\right)}{2}$

#### Explanation:

Use the chain rule (a lot of times!):

First issue: the cosecant function: $\frac{d}{\mathrm{dx}} \left(\csc \left(u\right)\right) = - \csc \left(u\right) \cot \left(u\right) \cdot u '$

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

Second issue: the square root: $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{d}{\mathrm{dx}} \left({u}^{\frac{1}{2}}\right) = \frac{1}{2} {u}^{- \frac{1}{2}} \cdot u '$

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

Here, recall that $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$.

$f ' \left(x\right) = - \csc \left(\sqrt{{e}^{x}}\right) \cot \left(\sqrt{{e}^{x}}\right) \cdot \frac{1}{2} {\left({e}^{x}\right)}^{- \frac{1}{2}} \cdot {e}^{x}$

To multiply ${\left({e}^{x}\right)}^{- \frac{1}{2}} \cdot {e}^{x}$, add the exponents to get ${\left({e}^{x}\right)}^{\frac{1}{2}} = \sqrt{{e}^{x}}$.

$f ' \left(x\right) = \frac{- \sqrt{{e}^{x}} \csc \left(\sqrt{{e}^{x}}\right) \cot \left(\sqrt{{e}^{x}}\right)}{2}$