Question

dy/dx of (sin x)^cos x​

Answer 1

Let y = ( sinx) ^ cos x

Taking log both sides,

log y = cos x log ( sinx)

Differentiating both sides now,

1/y ( dy/dx) = - sinx log( sinx) + cos x. cos x / sinx

dy/dx = y [ (- sin^2 x log( sinx) + cos^2 x) ÷ sin x]

dy/dx = sin x ^ ( cos x - 1) [- sin^2 x log( sinx) + cos^2 x ]