Question

What are the first and second derivatives of `g(x) = cosx^2 + e^(lnx^2)ln(x)`?

Answer 1

`g'(x) =-2xsin(x^2) + 2xln(x) + x`

Explanation:

This is a fairly standard chain and product rule problem.

The chain rule states that:
`d/dx f(g(x)) = f'(g(x))*g'(x)`

The product rule states that:
`d/dx f(x)*g(x) = f'(x)*g(x) + f(g)*g'(x)`

Combining these two, we can figure out `g'(x)` easily. But first let's note that:
`g(x) = cosx^2 + e^(lnx^2)ln(x) = cosx^2 + x^2ln(x)`

(Because `e^ln(x) = x` ). Now moving on to determining the derivative:
`g'(x) = -2xsin(x^2) + 2xln(x) + (x^2)/x`
`= -2xsin(x^2) + 2xln(x) + x`