Question

How do you differentiate `y = x^(cos x)`?

Answer 1

The answer is `=(1/xcosx-lnxsinx)x^(cosx)`

Explanation:

The function is

`y=x^(cosx)`

Taking logarithms on both sides

`lny=ln(x^(cosx))=cosxlnx`

Differentiating both sides

`(lny)'=(cosxlnx)'=(cosx)'lnx+cosx(lnx)'`

`1/ydy/dx=-sinx*lnx+cosx*1/x`

`dy/dx=(1/xcosx-lnxsinx)y`

`=(1/xcosx-lnxsinx)x^(cosx)`

Answer 2

`x^cosx(cosx/x-sinxlnx)`

Explanation:

We know that `a^b=e^(blna)`. Here, we say that `x^cosx=e^(cosxlnx)`.

According to the chain rule, `(df)/dx=(df)/(du)*(du)/dx`, where `u` is a function within `f`.

Here, `f=e^u` where `u=cosxlnx`, so we have:

`d/(du)e^u*d/dx(cosxlnx)`

`e^ud/dx(cosxlnx)`

According to the product rule, `(f*g)'=f'g+fg'`. Here, `f=cosx` and `g=lnx`, so we have:

`d/dxcosx*lnx+d/dxlnx*cosx`

`-sinxlnx+1/xcosx`

`cosx/x-sinxlnx`

So we have:

`e^u(cosx/x-sinxlnx)`, but as `u=cosxlnx`, we have:

`e^(cosxlnx)(cosx/x-sinxlnx)`, but remember that `e^(cosxlnx)=x^cosx`, so we have:

`x^cosx(cosx/x-sinxlnx)`