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

1 Answer
Jul 14, 2016

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