# How do you differentiate  y =-ln [ 3+ (9+x^2) / x]?

Dec 24, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{3 + \frac{9}{x} + x} \cdot \left(1 - \frac{9}{x} ^ 2\right)$

#### Explanation:

$y = - \ln \left(3 + \frac{9 + {x}^{2}}{x}\right) = - \ln \left(3 + \frac{9}{x} + x\right)$

Use the chain rule $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Let $y = - \ln \left(u\right)$ then $\frac{\mathrm{dy}}{\mathrm{du}} = - \frac{1}{u}$
And $u = 3 + \frac{9}{x} + x$ then $\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{9}{x} ^ 2 + 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{u} \cdot \left(1 - \frac{9}{x} ^ 2\right)$

Substitute $u = 3 + \frac{9}{x} + x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{3 + \frac{9}{x} + x} \cdot \left(1 - \frac{9}{x} ^ 2\right)$