Question

Find dy/dx of

y=x(lnx)^(sin^2x)

using log differentiation.

Answer 1

lny = ln(x(lnx)^(sin^2x))

Can someone help me distribute the ln on the right side I am having trouble

Answer 2

lny = ln x + (sin^2 x) ln (lnx)
y`/y = 1/x + sin2x (ln(lnx)) + sin^2x/(x lnx)

Answer 3

ln x + (sin^2 x) ln (lnx) can you try to explain this step more?