# What is the derivative of f(x)=(x-1)^2/ln(1/x)^2?

Dec 30, 2015

For the big picture, it's quotient rule. For both numerator and denominator, we'll need separate chain rules.

#### Explanation:

• Quotient rule: be $y = f \frac{x}{g} \left(x\right)$, then $y ' = \frac{f ' g - f g '}{g} ^ 2$

• Chain rule: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

Let's find the derivatives of both numerator and denominator separately then aggregate them up afterwards.

• Numerator: let's rename $u = x - 1$
$\frac{\delta y}{\delta x} = 2 u \left(1\right) = 2 x - 2$

• Denominator: let's rename $u = \frac{1}{x}$ and $v = \ln \left(u\right)$
$\frac{\delta y}{\delta x} = - \frac{2 v}{u x} = - \frac{2 \ln \left(\frac{1}{x}\right)}{\left(\frac{1}{x}\right) \left({x}^{2}\right)} = - \frac{2 \ln \left(\frac{1}{x}\right)}{x}$

Now, let's derivate the whole quotient:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(2 x - 2\right) \left(\ln {\left(\frac{1}{x}\right)}^{2}\right) + {\left(x - 1\right)}^{2} \frac{2 \ln \left(\frac{1}{x}\right)}{x}}{\ln} {\left(\frac{1}{x}\right)}^{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\left(2 x\right) \left(x - 1\right) \left(\ln {\left(\frac{1}{x}\right)}^{2}\right) + 2 {\left(x - 1\right)}^{2} \ln \left(\frac{1}{x}\right)}{x}}{\ln} {\left(\frac{1}{x}\right)}^{4}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(x - 1\right) \left(\ln \left(\frac{1}{x}\right)\right) \left(2 x \ln \left(\frac{1}{x}\right) + 2 \left(x - 1\right)\right)}{x \ln {\left(\frac{1}{x}\right)}^{4}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \left(x - 1\right) \left(x \ln \left(\frac{1}{x}\right) + x - 1\right)}{x \ln {\left(\frac{1}{x}\right)}^{3}}$