Question

How do you differentiate `y=13^sinx`?

Answer 1

`13^(sin(x)) * cos(x) * ln(13)`
using the Chain

Explanation:

start by Changing the look of the equation a little bit
Realize that 13 is equal to `e^(ln(13))` as `ln` which is the natural log is the log base `e`

`therefore` the equation becomes `(e^(ln(13)))^sin(x)`
and according to index rules, it is equal to `e^(ln(13)*sin(x)`

now let `f(x) = e^x`
and `g(x) = ln(13) * sin(x)`
so `y = f(g(x))`

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

`= d/dx y = f'(g(x))* g'(x)`

therefore on evalation, the answer becomes

`e^(ln(13)*sin(x))*cos(x)*ln(13)`

`therefore` `(e^(ln13))^(sin(x)) * cos(x) * ln(13)`

= `13^(sin(x)) * cos(x) * ln(13)`