Question

Question #cab31

Answer 1

`-cotx. lnx -1/x`

Explanation:

`y = ln (x^sinx)`
`y= sinx lnx`

Differentiating wrt to `cosx`:

Using product rule:

`dy/(d(cosx)) = (dsinx) /(d (cosx)). lnx + (dlnx)/(d(cosx)). sinx`

Differentiating `sinx` wrt to `cosx` is equivalent to differenting both wrt to `x`
i.e `((dsinx)/dx )/((dcosx)/dx)`

`dy/(d(cosx)) = cosx /-sinx. lnx + (1/x)/-sinx. sinx`

`dy/(d(cosx)) = -cotx. lnx - 1/x`