Question

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

Answer 1

`dy/dx = x^cosx(-sinxlnx + cosx/x)`

Explanation:

`y = x^cosx`

Take the natural logarithm of both sides.

`lny = ln(x^cosx)`

Use the logarithm law for powers, which states that `loga^n = nloga`

`lny = cosxlnx`

Use the product rule to differentiate the right hand side. `d/dx(cosx) = -sinx` and `d/dx(lnx)`.

`1/y(dy/dx) = -sinx(lnx) + cosx(1/x)`

`1/y(dy/dx) = -sinxlnx + cosx/x`

`dy/dx = (-sinxlnx + cosx/x)/(1/y)`

`dy/dx = x^cosx(-sinxlnx + cosx/x)`

Hopefully this helps!

Answer 2

`d/(dx) x^(cosx) = x^(cosx) (cosx/x -sinx lnx )`

Explanation:

You can write:

`x^(cosx) = (e^lnx)^(cosx) = e^(lnxcosx)`

so:

`d/(dx) x^(cosx) = d/(dx) (e^(lnxcosx))`

using the chain rule:

`d/(dx) x^(cosx) = e^(lnxcosx) d/(dx) (lnxcosx)`

then the product rule:

`d/(dx) x^(cosx) = x^(cosx) (cosx/x -sinx lnx )`