# What is the derivative of f(x)=(lnx)^2/x^2?

Oct 19, 2016

$f ' \left(x\right) = \frac{2 \ln x \left(1 - \ln x\right)}{x} ^ 3$

#### Explanation:

We can rewrite $f \left(x\right) = {\left(\ln x\right)}^{2} / {x}^{2}$ as $f \left(x\right) = {\left(\ln \frac{x}{x}\right)}^{2}$

Now, let $u = \ln \frac{x}{x}$ Then using the product rule:

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$

So $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{x \frac{d}{\mathrm{dx}} \left(\ln x\right) - \left(\ln x\right) \frac{d}{\mathrm{dx}} \left(x\right)}{x} ^ 2$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = \frac{x \frac{1}{x} - \left(\ln x\right) 1}{x} ^ 2$
$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1 - \ln x}{x} ^ 2$

And using the chain rule we now have:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left\{{\left(\ln \frac{x}{x}\right)}^{2}\right\}$
$\therefore f ' \left(x\right) = \frac{d}{\mathrm{dx}} {u}^{2}$
$\therefore f ' \left(x\right) = \frac{d}{\mathrm{du}} {u}^{2} \times \frac{\mathrm{du}}{\mathrm{dx}}$
$\therefore f ' \left(x\right) = 2 u \frac{\mathrm{du}}{\mathrm{dx}}$

And substituting $u$ and $\frac{\mathrm{du}}{\mathrm{dx}}$ gives:
$\therefore f ' \left(x\right) = 2 \ln \frac{x}{x} \times \frac{1 - \ln x}{x} ^ 2$
$\therefore f ' \left(x\right) = \frac{2 \ln x \left(1 - \ln x\right)}{x} ^ 3$