# How do you differentiate y=(1+lnx)/(x^2-lnx) using the quotient rule?

Aug 28, 2017

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

#### Explanation:

$y = \frac{1 + \ln x}{{x}^{2} - \ln x}$

$\frac{d}{\mathrm{dx}} \left(\frac{p}{q}\right) = \frac{q p ' - p q '}{q} ^ 2$

$\frac{d}{\mathrm{dx}} \left(1 + \ln x\right) = \frac{1}{x}$

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

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

Expanding $\frac{\mathrm{dy}}{\mathrm{dx}}$ and multiplying by $\frac{x}{x}$ gives

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