Derivative of (sin(x))^x

What are the steps used to obtain the derivative of (sin(x))^x? I know it's (sin(x))^x [xcot(x) + ln(sinx)] however I don't know how to get there.

Edit:

y = sin(x)^x
ln(y) = x ln(sin(x))

Now differentiate implicitly.

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or: (d/dx)sin(x)^x = (d/dx)e^xln(sin(x)) = (ln(sin(x))+xcosx/sinx)sin(x)^x, that's what you got.

1st
lny=ln((sinx)^x)
lny=xln(sinx)
then take the derivative and use product and chain rule ;)

dy/dx(1/y)=x(1/sinx)cosx + ln(sinx)
then simplfy

dy/dx(1/y) = xcotx+lnsinx
dy/dx=y[xcotx+lnsinx]

plug in for y

dy/dx= (sinx)^x[xcotx+lnsinx]

message me if you need any explanations.

dw soz

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Thats a fun one to solve.

how do you integrate (sin(x))^x?

For the second derivative, I got

(((sin(x))^x)*(ln(sinx)+x/tanx))*(ln(sinx)+x*cotx)+((sin(x))^x)*(cotx-cotx/((sin(x))^2)

but got no way of checking

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uh guys, are you even allowed to do ln(sinx), i mean sinx isn't always positive right?

HallsofIvy
Homework Helper
Then do it for $0\le x\le \pi$ and use $sin(x+ n\pi)= (-1)^n sin(x)$ for x outside that range.

kk, so how do i differentiate (-1)^x?

my question is: how do do ln((-1)^n*sinx)?

if I take -Pi<x<0 then I can write:

sin(x)^x=(-1)^x*(-sin(x))^x
so sin(x)^x=(-1)^x*exp(x*ln(-sin(x))) but then I'm stuck in the differentiation because of the (-1)^x

HallsofIvy
The problem is not just differentiating $(-1)^x$. How are you defining $(-1)^x$?