Question

find f'(x)

(sinx)^cosx

Answer 1

See the generalized power rule here:

http://en.wikipedia.org/wiki/Table_of_derivatives

Answer 2

Use the chain rule for differentiating a function of a function.

Let cosx = v(x)

d/dx[(sinx)^v(x)]= d(sinx)^v/dv * dv/dx
= d/dv[sqrt(1-v^2)]/dv * dv/dx
= (-2)(1/2)/[sqrt(1-v^2)] * (-sin x)
= (-1)/sin x * (-sin x) = 1

A very interesting result!

Answer 3

My answer looks rather suspicious, although I verified it at x = 0 and x = 0.5. I hope that other teachers will respond.

Answer 4

I would do it this way:
take ln of both sides

lny = ln(cosx^sinx)
= sinx(ln(cosx))
(dy/dx)/y = sinx(cosx/sinx) + (-sinx)(ln(sinx)

dy/dx = (cosx)^sinx[sinx(cosx/sinx) + (-sinx)(ln(sinx)]

don't know how much simplication is needed.

Answer 5

Got my sines and cosines mixed up as I copied from paper.

try again:

lny = ln(sinx^cosx)
= cosx(ln(sinx))

(dy/dx)/y = cosx(cosx/sinx) + (-sinx)(ln(sinx))
dy/dx = dy/dx = (sinx)^cosx[cosx(cosx/sinx) + (-sinx)(ln(sinx)]