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

1 Answer
Dec 23, 2017

13^(sin(x)) * cos(x) * ln(13)13sin(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))eln(13) as lnln which is the natural log is the log base ee

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)