# How do you use the quotient rule to differentiate (2lnx^3) / sqrt(2-x^2)?

Mar 18, 2018

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 6 \left[\frac{2 - {x}^{2} + {x}^{2} \ln x}{x {\left(2 - {x}^{2}\right)}^{\frac{3}{2}}}\right]$

#### Explanation:

Here, $2 \ln {x}^{3} = 2 \left(3 \ln x\right) = 6 \ln x$

So, let us take

$y = \frac{6 \ln x}{\sqrt{2 - {x}^{2}}}$

Now, $\textcolor{red}{\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot {u}^{'} - u \cdot {v}^{'}}{v} ^ 2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 6 \left[\frac{\sqrt{2 - {x}^{2}} \cdot \frac{1}{x} - \ln x \cdot \frac{1}{\cancel{2} \left(\sqrt{2 - {x}^{2}}\right)} \left(- \cancel{2} x\right)}{{\left(\sqrt{2 - {x}^{2}}\right)}^{2}}\right]$

$= 6 \left[\frac{\frac{\sqrt{2 - {x}^{2}}}{x} + \frac{x \ln x}{\sqrt{2 - {x}^{2}}}}{2 - {x}^{2}}\right]$

$= 6 \left[\frac{\left(2 - {x}^{2}\right) + \left({x}^{2} \ln x\right)}{x \sqrt{2 - {x}^{2}} \left(2 - {x}^{2}\right)}\right]$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = 6 \left[\frac{2 - {x}^{2} + {x}^{2} \ln x}{x {\left(2 - {x}^{2}\right)}^{\frac{3}{2}}}\right]$