Question

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

Answer 1

`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.