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

Feb 16, 2016

$f ' \left(x\right) = - \frac{4}{x}$

#### Explanation:

Given equation is $f \left(x\right) = 2 \ln {\left(\frac{1}{x}\right)}^{2}$
From standard logarithm equations, we know that $\ln {x}^{n} = n \ln x$
So by shifting the indices, we get
$f \left(x\right) = - 4 \ln x$

We now see that the differentiation need not even need chain rule, hence the given derivative becomes $f ' \left(x\right) = - \frac{4}{x}$

But, if still only chain rule must be applied, then differentiate it nonetheless as such
$f ' \left(x\right) = \frac{2}{\frac{1}{x} ^ 2} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x} ^ 2\right) = 2 {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({x}^{-} 2\right) = 2 {x}^{2} \cdot - 2 {x}^{- 2 - 1}$

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