Derivative of (sin(x))^x

Edit:

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

Now differentiate implicitly.

cookiemonster

 

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

 

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

 

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

 

Then do it for [itex]0\le x\le \pi[/itex] and use [itex]sin(x+ n\pi)= (-1)^n sin(x)[/itex] for x outside that range.

 

kk, so how do i differentiate (-1)^x?
edit: sorry i misread your post

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

 

The problem is not just differentiating [itex](-1)^x[/itex]. How are you defining [itex](-1)^x[/itex]?

 

I dunno, hence my question: how do you guys manage to differentiate sinx^x?

When I try to differentiate it using the suggested method which I get stuck because of the (-1)^x for values of x on the intervals of the form [2kPi-Pi,2kPi].

This is frustrating I know, everywhere I look people use the same method, but to me there is something missing , or maybe there is something wrong with my thinking :(