# The derivative of natural log?

## $y = - \ln \left(\frac{3}{2} x\right)$

Jun 18, 2018

$y ' = - \frac{1}{x}$

#### Explanation:

We have: $y = - \ln \left(\frac{3}{2} x\right)$

We must use the chain rule to differentiate $y$.

Let $u = \frac{3}{2} x R i g h t a r r o w u ' = \frac{3}{2}$ and $v = - \ln \left(u\right) R i g h t a r r o w v ' = - \frac{1}{u}$:

$R i g h t a r r o w y ' = u ' \cdot v '$

$R i g h t a r r o w y ' = \frac{3}{2} \cdot - \frac{1}{\frac{3}{2} x}$

$R i g h t a r r o w y ' = \frac{3}{2} \cdot - \frac{2}{3 x}$

$\therefore y ' = - \frac{1}{x}$

Jun 18, 2018

$- \frac{1}{x}$

#### Explanation:

Remember the form:

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

$y = - \ln \left(\frac{3 x}{2}\right)$

$y = \ln \left(\frac{2}{3 x}\right)$

$\setminus \therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2}{3 {x}^{2}} \setminus \div \frac{2}{3 x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x}$

Hope that makes sense!