Question

Find the derivative of the following function: (x^2)^sinx.

Answer 1

if y = u^v, where u and v are functions of x,

y' = v u^(v-1) u' + ln(u) u^v v'

If you look carefully, you will see that
y=u^n and y=a^v are just special cases, where u' or v' = 0.

Anyway, just plug in

u = x^2
v = sinx

and you get

(2x sinx)*(x^2)(sinx-1) + ln(x^2)(cosx)*(x^2)^sinx

or

(x^2)(sinx) (2x*sinx/x^2 + 2lnx*cosx)
= 2x^(2sinx) (sinx/x + lnx*cosx)