AP Calculus AB Question

find f'(x) of f(x)=x^cotx

f(x) = x^(cot(x))
ln(f(x)) = cot(x) * ln(x)
f'(x) / f(x) = cot(x) / x - csc(x)^2 * ln(x)
f'(x) = f(x) * (cot(x) / x - csc(x)^2 * ln(x))
f'(x) = (x^cot(x)) * (cot(x) / x - csc(x)^2 * ln(x))

Possible derivation:
d/dx(f(x)) = d/dx(x^(cot(x)))
| The derivative of f(x) is f'(x):
= | f'(x) = d/dx(x^(cot(x)))
| Use the chain rule, d/dx(x^(cot(x))) = ( du^v)/( du) ( du)/( dx)+( du^v)/( dv) ( dv)/( dx), where u = x, v = cot(x) and ( du^v)/( du) = u^(-1+v) v, ( du^v)/( dv) = u^v log(u):
= | f'(x) = cot(x) x^(cot(x)-1) (d/dx(x))+log(x) x^(cot(x)) (d/dx(cot(x)))
| The derivative of x is 1:
= | f'(x) = log(x) x^(cot(x)) (d/dx(cot(x)))+cot(x) x^(cot(x)-1)
| The derivative of cot(x) is -csc^2(x):
= | f'(x) = cot(x) x^(cot(x)-1)+log(x) x^(cot(x)) -csc^2(x)

Take the natural log, exponentiate and simplify:
e^[ln(x^cotx)]
e^[cotx*ln(x)]

Differentiate using the chain rule; in general,
d(e^u)/dx = e^u * du/dx

d([cotx*ln(x)])/dx = [csc^2(x)]*ln(x) + (cotx)/x