# How do you differentiate y=13^sinx?

##### 1 Answer
Dec 23, 2017

${13}^{\sin \left(x\right)} \cdot \cos \left(x\right) \cdot \ln \left(13\right)$
using the Chain

#### Explanation:

start by Changing the look of the equation a little bit
Realize that 13 is equal to ${e}^{\ln \left(13\right)}$ as $\ln$ which is the natural log is the log base $e$

$\therefore$ the equation becomes ${\left({e}^{\ln \left(13\right)}\right)}^{\sin} \left(x\right)$
and according to index rules, it is equal to e^(ln(13)*sin(x)

now let $f \left(x\right) = {e}^{x}$
and $g \left(x\right) = \ln \left(13\right) \cdot \sin \left(x\right)$
so $y = f \left(g \left(x\right)\right)$

https://en.wikipedia.org/wiki/Chain_rule
using the chain rule, the derivative of $y$ is equal to
the derivative of the outside function evaluated at the inside function multiplied by the Derivative of the inside function

$= \frac{d}{\mathrm{dx}} y = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

therefore on evalation, the answer becomes

${e}^{\ln \left(13\right) \cdot \sin \left(x\right)} \cdot \cos \left(x\right) \cdot \ln \left(13\right)$

$\therefore$ ${\left({e}^{\ln 13}\right)}^{\sin \left(x\right)} \cdot \cos \left(x\right) \cdot \ln \left(13\right)$

= ${13}^{\sin \left(x\right)} \cdot \cos \left(x\right) \cdot \ln \left(13\right)$