# How do you implicitly differentiate -y^2=e^(2x-4y)-3x ?

Jun 29, 2016

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 - 2 {e}^{2 x - 4 y}}{4 \left\{2 y - {e}^{2 x - 4 y}\right\}} .$

OR

.dy/dx={3-6x+2y^2}/{4(2y-3x+y^2}.

#### Explanation:

We write the eqn. of the curve as $: 3 x - {y}^{2} = {e}^{2 x - 4 y} \ldots \ldots \ldots . \left(1\right)$

Diff.ing both sides w.r.t. x, $\frac{d}{\mathrm{dx}} \left(3 x - 4 {y}^{2}\right) = \frac{d}{\mathrm{dx}} {e}^{2 x - 4 y} .$

$\therefore \frac{d}{\mathrm{dx}} 3 x - \frac{d}{\mathrm{dx}} 4 {y}^{2} = {e}^{2 x - 4 y} \cdot \frac{d}{\mathrm{dx}} \left(2 x - 4 y\right) .$
$\therefore 3 - 4 \frac{d}{\mathrm{dy}} \left({y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x - 4 y} \left(\frac{d}{\mathrm{dx}} 2 x - \frac{d}{\mathrm{dx}} 4 y\right)$......[Chain Rule]
$\therefore 3 - 4 \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{2 x - 4 y} \left\{2 - 4 \frac{\mathrm{dy}}{\mathrm{dx}}\right\} .$
$\therefore 3 - 2 {e}^{2 x - 4 y} = 8 y \frac{\mathrm{dy}}{\mathrm{dx}} - 4 {e}^{2 x - 4 y} \frac{\mathrm{dy}}{\mathrm{dx}} = 4 \left\{2 y - {e}^{2 x - 4 y}\right\} \frac{\mathrm{dy}}{\mathrm{dx}} .$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 - 2 {e}^{2 x - 4 y}}{4 \left\{2 y - {e}^{2 x - 4 y}\right\}} .$

This is fairly a good answer, but, we can further simplify it, replacing ${e}^{2 x - 4 y} ,$ by $3 x - {y}^{2}$ [bcz. of $\left(1\right)$], as under :-

.dy/dx={3-2(3x-y^2)}/[4{2y-(3x-y^2)}]={3-6x+2y^2}/{4(2y-3x+y^2}.