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#