What is the derivative of (sin x)^x for all x on the interval [0,pi/2]?

Question

I'm not sure how to proceed doing this.

Answer 1

Take the log of both sides:

y = (sin x)^x

ln y = ln (sin x)^x

ln y = x ln(sin x)

Take the deriv. of both sides:

1/y(dy/dx) = x*1/(sinx)*cos x + ln(sin x)

Multiply both sides by y:

dy/dx = y[x*cos(x)/sin(x) + ln(sin x)]

Then replace y by (sin x)^x

dy/dx = (sin x)^x[x*cot(x) + ln(sin x)]

Answer 2

This makes no sense. And I agree with the previous person. How can you evaluate a derivate over an interval. Are you sure this not an integral?

I think I have seen then integral of this, and basically you have to look it up in a table of the top of my head it is like synch (x). Please edit your question, and I will check back later.

Answer 3

it is a chain rule you have to take the derivative of the inside then the outside of the equation. However since you are giving an interval, are you sure that this isn't an integral problem?