# How do you differentiate f(x)= sqrt (ln(1/x)/x^2?

Dec 29, 2015

Using chain and quotient rule.

#### Explanation:

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

• Quotient rule states that for $y = f \frac{x}{g} \left(x\right)$, $y ' = \frac{f ' g - f g '}{g} ^ 2$

We can go renaming parts of the expression in order to make up the chain.

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

$\left(\mathrm{df} \left(x\right)\right) = \left(\mathrm{dx}\right) = \frac{1}{2 {u}^{\frac{1}{2}}} \cdot \frac{\left(\frac{1}{v}\right) \left(- \frac{1}{x} ^ 2\right) \left({x}^{2}\right) - \ln \left(\frac{1}{x}\right) 2 x}{x} ^ 4$

Substituting $v$, then $u$ and aggregating what's possible, so far

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{1}{2 {u}^{\frac{1}{2}}} \cdot \frac{\left(\frac{1}{\frac{1}{x}}\right) \left(- \frac{1}{\cancel{{x}^{2}}}\right) \left(\cancel{{x}^{2}}\right) - \ln \left(\frac{1}{x}\right) 2 x}{x} ^ 4$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{1}{2 {\left(\frac{\ln \left(\frac{1}{x}\right)}{x} ^ 2\right)}^{\frac{1}{2}}} \cdot \frac{\left(x\right) \left(- \frac{1}{\cancel{{x}^{2}}}\right) \left(\cancel{{x}^{2}}\right) - \ln \left(\frac{1}{x}\right) 2 x}{x} ^ 4$

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{- x - 2 x \ln \left(\frac{1}{x}\right)}{2 {x}^{4} \sqrt{\frac{\ln \left(\frac{1}{x}\right)}{x} ^ 2}} = \textcolor{g r e e n}{\frac{- 1 - 2 \ln \left(\frac{1}{x}\right)}{2 {x}^{3} \sqrt{\frac{\ln \left(\frac{1}{x}\right)}{x} ^ 2}}}$