# Hoe do you differentiate f(x)=ln(1/x^2) - ln(1/x^3) ?

Nov 12, 2015

By using logarithmic properties to create one natural log, then using $d \left(\ln \left(u\right)\right) = \frac{d \left(u\right)}{u}$

#### Explanation:

The derivative of a log $\ln \left(u\right)$ is the derivative of $u$ divided by $u$. ($\frac{\mathrm{du}}{u}$)

For the sake of ease, I'm going to change the equation to $f \left(x\right) = \ln \left({x}^{-} 2\right) - \ln \left({x}^{-} 3\right)$ so that I may use the power rule instead of going through the entirety of the quotient rule. Remember it's better to work smart, not hard.

So we're going to go through each term in the equation and take the derivative of it.

The derivative of $\ln \left({x}^{-} 2\right)$ is $\frac{- 2 {x}^{-} 3}{x} ^ - 2$ which simplifies to $- \frac{2}{x}$.

The derivative of $\ln \left({x}^{-} 3\right)$ is $\frac{- 3 {x}^{-} 4}{x} ^ - 3$ which simplifies to $- \frac{3}{x}$.

Now that we have our derivatives of each term, let's go back to our equation.

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

Now, that is the correct answer, but after finishing this problem, I realized that I worked hard, and not smart. This problem becomes a whole lot easier when we implement the log rules, specifically $\ln \left(a\right) - \ln \left(b\right) = \ln \left(\frac{a}{b}\right)$.

So implementing this...

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

And very simply, the derivative of $\ln \left(x\right)$ is $\left(\frac{1}{x}\right)$.