How do you differentiate #f(x)=2ln(1/x)^2# using the chain rule?

1 Answer
Feb 16, 2016

Answer:

#f'(x)=-4/x#

Explanation:

Given equation is #f(x)=2ln(1/x)^2#
From standard logarithm equations, we know that #lnx^n=nlnx#
So by shifting the indices, we get
#f(x)=-4lnx#

We now see that the differentiation need not even need chain rule, hence the given derivative becomes #f'(x)=-4/x#

But, if still only chain rule must be applied, then differentiate it nonetheless as such
#f'(x)=2/(1/x^2)*d/dx(1/x^2)=2x^2*d/dx(x^-2)=2x^2*-2x^(-2-1)#

So in the end, simplification ends to the same method as given above.