Strange Differentiations (1 Viewer)

[Inexorably Me]

Q: Find d/dx of ln [(Sinx)^x]
A: lnSinx + xcotx

Hmm, i tried using substitution for (Sinx)^x as a^x but i keep getting +cosx instead of xcotx...Please help >_<! Your efforts are much appreciated. <3

 

[AlexJB]

And then it's normal chain rule.
f'(x) = ln(sinx)(1) + cotx(x)
= ln(sinx) + xcotx