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

##### 1 Answer
Jan 7, 2017

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

#### Explanation:

$y = {e}^{x + 2 y}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + 2 y} \frac{d}{\mathrm{dx}} \left(x + 2 y\right)$

$= {e}^{x + 2 y} \left\{1 + 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right\} = {e}^{x + 2 y} + 2 {e}^{x + 2 y} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {e}^{x + 2 y} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + 2 y}$

$\therefore \left\{1 - 2 {e}^{x + 2 y}\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + 2 y}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x + 2 y} / \left(1 - 2 {e}^{x + 2 y}\right) = \frac{y}{1 - 2 y}$

As we have been asked to differentiate implicitly, we proceeded as above. Otherwise , we can use the following method :

$y = {e}^{x + 2 y} \Rightarrow \ln y = x + 2 y \Rightarrow \ln y - 2 y = x$

Now diff.ing both sides w.r.t. $y , \frac{1}{y} - 2 = \frac{\mathrm{dx}}{\mathrm{dy}} , \mathmr{and} \frac{1 - 2 y}{y} = \frac{\mathrm{dx}}{\mathrm{dy}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = \frac{y}{1 - 2 y}$

Enjoy Maths.!