Can anyone help with the Logarithmic differentiation of

Question

(sin(x))^x?

Answer 1

The expression is (sin(x))^x

Rewrite the expression, let y = (sin x)^x

Take log each side

log y = log[ (sin x)^x ] = x log (sin x)

Differentiate and apply derivative formula: (uv)' = uv' + vu'.

y' / y = x [ {1/(sin x)} * cos x ] + log(sin x)

= x cot x + log (sin x)

y' = y { x cot x + log (sin x) }

= (sin x)^x { x cot x + log (sin x) }

The solution is (sin x)^x { x cot x + log (sin x) }.