Question

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

Answer 1

`d/dx(x^sinx)=x^sinx(cosxlogx+sinx/x)` where `logx` is the natural logarithm.

Explanation:

Let `y=x^sinx`. We will use logarithmic differentiation to calculate `dy/dx`.
`logy=log(x^sinx)`
`logy=sinxlogx`
Now, differentiating both sides with respect to `x`, we have
`d/dx(logy)=d/dx(sinxlogx)`

To compute the left side, remember the chain rule:
`y=f(g(x))iffdy/dx=f'(g(x))g'(x)`. Letting `f(x)=logx` and `g(x)=y`, the left side becomes `d/dx(logy)=1/ydy/dx`.
`:.1/ydy/dx=d/dx(sinxlogx)`

To differentiate `sinxlogx`, remember the product rule:
`y=f(x)g(x)iffdy/dx=f'(x)g(x)+g'(x)f(x)`
`:.1/ydy/dx=cosxlogx+sinx/x`

Multiplying through by `y`, we have
`dy/dx=y(cosxlogx+sinx/x)`
Remembering that we defined `y=x^sinx`, we get
`dy/dx=x^sinx(cosxlogx+sinx/x)`