Derivative of $ y = \frac{1}{\ln^{2}x} $

Question

I am supposed to find the derivative of $ y = \frac{1}{\ln^{2}x} $. How would you calculate it? My first step was to do this: $\frac{-1\ast \ln^{2}x}{(\ln^{2}x)^{2}}$. How would you continue? I don't know what to do with $ \ln^{2}x $. Thanks

Answer 1

You are differentiating $$ \frac{1}{(\ln x)^2} = f(g(x)), $$ where $$ f(x) = \frac{1}{x^2}, \quad g(x) = \ln x. $$ Since $$ f'(x) = -\frac{2}{x^3}, \quad g'(x) = \frac{1}{x}, $$ the chain rule therefore gives $$ \frac{\mathrm d}{\mathrm dx} \frac{1}{(\ln x)^2} = \frac{\mathrm d}{\mathrm dx} f(g(x)) = f'(g(x))g'(x) = -\frac{2}{(\ln x)^3} \frac{1}{x}. $$ You could do it even more directly: $$ \frac{\mathrm d}{\mathrm dx} \frac{1}{(\ln x)^2} = -\frac{2}{(\ln x)^3}\frac{\mathrm d}{\mathrm dx}\ln x = -\frac{2}{x(\ln x)^3}. $$

Answer 2

Hint:

Use the chain rule, and remember that $$\biggl(\frac1{x^2}\biggr)'=-\frac 2{x^3},\enspace\text{more generally: }\quad\biggl(\frac1{x^n}\biggr)'=-\frac n{x^{n+1}}$$

Answer 3

A rule that always hold is $(u^n)' = nu'u^{n-1}$.

With $u = ln$ we get $\forall x\in \mathbb{R}^*_+, u^{-2}(x) = \frac{-2}{x}ln^{-3}(x)$.