Jtdylktuy

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: $frac{-1ast ln^{2}x}{(ln^{2}x)^{2}}$. How would you continue? I don't know what to do with $ ln^{2}x $. Thanks

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:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$

Hint:

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

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

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