Question

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

Answer 1

`dy/dx = (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)`

Take Natural logarithms:

`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}`