Question

use logarithmic differentiation to find:

y = (sinx)^lnx

Answer 1

y = (sin(x))^(ln(x))

Taking the natural log of both sides:

ln(y) = ln(x)*ln(sin(x))

Differentiating both sides:

y'/y = ln(x)*cot(x) + ln(sin(x))/x

y' = [(sin(x))^(ln(x))] * [ln(x)*cot(x) + ln(sin(x))/X]