# How do you find dy/dx by implicit differentiation given y^2=ln(2x+3y)?

Feb 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{6 {y}^{2} + 4 x y - 3} .$

#### Explanation:

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

$\therefore \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left\{\ln \left(2 x + 3 y\right)\right\} .$

Using the Chain Rule, we have,

$\frac{d}{\mathrm{dy}} \left({y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 x + 3 y} \frac{d}{\mathrm{dx}} \left(2 x + 3 y\right) .$

$\therefore 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 x + 3 y} \left\{2 + 3 \frac{\mathrm{dy}}{\mathrm{dx}}\right\} .$

$= \frac{2}{2 x + 3 y} + \frac{3}{2 x + 3 y} \frac{\mathrm{dy}}{\mathrm{dx}} .$

$\therefore \left\{2 y - \frac{3}{2 x + 3 y}\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{2 x + 3 y} .$

$\therefore , \mathmr{if} \left(2 x + 3 y\right) \ne 0 , \left\{\frac{4 x y + 6 {y}^{2} - 3}{\cancel{2 x + 3 y}}\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{\cancel{2 x + 3 y}} .$

As, $\left(2 x + 3 y\right) = 0$ makes $\ln \left(2 x + 3 y\right)$ undefined, we have,

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2}{6 {y}^{2} + 4 x y - 3} .$