# How do you differentiate e^y = (tanx)^sinx?

Feb 24, 2017

$y ' = \cos x \ln \left(\tan x\right) + \sec x$

#### Explanation:

Take the natural logarithm of both sides.

$\ln \left({e}^{y}\right) = \ln {\left(\tan x\right)}^{\sin} x$

$y \ln e = \sin x \ln \left(\tan x\right)$

$y = \sin x \ln \left(\tan x\right)$

This is now a normal function explicitly defined by $y$. The derivative can be found by using the product and chain rules.

Let $y = g \left(x\right) h \left(x\right)$, with $g \left(x\right) = \sin x$ and $h \left(x\right) = \ln \left(\tan x\right)$. The derivative of $g \left(x\right)$, by first principles, in $g ' \left(x\right) = \cos x$.

However, we must use the chain rule to find $h ' \left(x\right)$. We let $y = \ln u$ and $u = \tan x$. Then $\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dx}} = {\sec}^{2} x$. This means that $h ' \left(x\right) = \frac{1}{u} \cdot {\sec}^{2} x = {\sec}^{2} \frac{x}{\tan} x = \frac{\frac{1}{\cos} ^ 2 x}{\sin \frac{x}{\cos} x} = \sec x \csc x$

We now apply the product rule, which states that $y ' = g ' \left(x\right) h \left(x\right) + h ' \left(x\right) g \left(x\right)$.

$y ' = \cos x \left(\ln \left(\tan x\right)\right) + \sin x \left(\sec x \csc x\right)$

$y ' = \cos x \ln \left(\tan x\right) + \sec x$

Hopefully this helps!