How do you differentiate y=(sinx)^lnxy=(sinx)lnx?

1 Answer
Feb 21, 2017

dy/dx = (sinx)^(lnx){(lnx)(cotx) + (ln(sinx))/x} dydx=(sinx)lnx{(lnx)(cotx)+ln(sinx)x}

Explanation:

Generally when dealing with a variable exponent it is easier to differentiate (and understand) by taking natural logarithms (to remove the exponent) and differentiating implicitly:

We have:

y = (sinx)^(lnx) y=(sinx)lnx

Take Natural logarithms:

lny = ln{(sinx)^(lnx)} lny=ln{(sinx)lnx}
\ \ \ \ \ \= (lnx)(ln(sinx)) \ \ \ (rule of logs)

Differentiate wrt x (LHS implicitly; RHS product rule with chain rule):

\ 1/ydy/dx = (lnx)(1/sinx*cosx) + (1/x)(ln(sinx))
\ \ \ \ \ \ \ \ \ \ \= (lnx)(cotx) + (ln(sinx))/x

:. dy/dx = y{(lnx)(cotx) + (ln(sinx))/x}
\ \ \ \ \ \ \ \ \ \ = (sinx)^(lnx){(lnx)(cotx) + (ln(sinx))/x}