Question

What is the derivative of `y = (sin x)^(cos x)`?

Answer 1

`dy/dx = (sinx)^cosx(-sinxln(sinx) + cosxcotx)`

Explanation:

Take the natural logarithm of both sides.

`lny = ln(sinx)^cosx`

Use the rule `loga^n = nloga` to simplify:

`lny = cosxln(sinx)`

Use the implicit differentiation as well as the product and chain rules to differentiate.

`d/dx(lnsinx) = 1/sinx * cosx = cosx/sinx = cotx`

Now to the relation as a whole:

`1/y(dy/dx) = -sinx(ln(sinx)) + cosxcotx`

Solve for `dy/dx`:

`dy/dx = y(-sinxln(sinx) + cosxcotx)`

`dy/dx = (sinx)^cosx(-sinxln(sinx) + cosxcotx)`

Hopefully this helps!