# Hoe do you differentiate f(x)=xln(1/x^2) ?

Dec 13, 2015

Use some algebra to change the form of the function, and then apply the product rule to find
$f ' \left(x\right) = \ln \left(\frac{1}{x} ^ 2\right) - 2$

#### Explanation:

If we were to differentiate directly, we would need to use the product rule, the chain rule, and the quotient rule. However, with some basic algebra, we can make it so all we need is the product rule.

First, the algebra:

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

Now, given
$\frac{d}{\mathrm{dx}} - 2 x = - 2$
and
$\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x}$

we use the product rule to obtain

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- 2 x\right) \ln \left(x\right)$

$= \left(- 2 x\right) \left(\frac{d}{\mathrm{dx}} \ln \left(x\right)\right) + \left(\frac{d}{\mathrm{dx}} - 2 x\right) \left(\ln \left(x\right)\right)$

$= - 2 x \left(\frac{1}{x}\right) + \left(- 2\right) \ln \left(x\right)$

$= - 2 \ln \left(x\right) - 2$

$= \ln \left({x}^{- 2}\right) - 2$

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