# Question #82351

Feb 24, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} \left(\ln \left(\sin x\right) + \cot x\right)$

#### Explanation:

To find the derivative of this expression, we'll use the product rule, which states:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$
Where $u$ and $v$ are functions of some variable (usually $x$).

We can see that the problem satisfies the two basic requirements of the product rule: both ${e}^{x}$ and $\ln \left(\sin x\right)$ are functions of $x$, and they're being multiplied together. So we can let $u = {e}^{x}$ and $v = \ln \left(\sin x\right)$ and apply the rule. First, we'll have to find the derivatives of these two functions:
$\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$

$\frac{d}{\mathrm{dx}} \left(\ln \left(\sin x\right)\right) = \cos x \cdot \frac{1}{\sin} x = \cos \frac{x}{\sin} x = \cot x \to$using chain rule and $\ln x$ rule

Now we have $u$ $\left({e}^{x}\right)$, $u '$ $\left({e}^{x}\right)$, $v$ $\left(\ln \left(\sin x\right)\right)$, and $v '$ $\left(\cot x\right)$. All that is left is to plug these guys into the formula and simplify:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \left({e}^{x}\right) \left(\ln \left(\sin x\right)\right) + \left({e}^{x}\right) \left(\cot x\right)$
$= {e}^{x} \left(\ln \left(\sin x\right) + \cot x\right)$

Remember that $u$ and $v$ don't have to be functions of $x$; they can be functions of anything as long as it's the same variable (as in $3 {z}^{2} \cos z$).