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

##### 1 Answer
Mar 31, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 - {e}^{y} - y}{x {e}^{y} + x + 8 - 4 y} .$

#### Explanation:

To avoid Quotient Rule for Diffn., we first, rewrite the given eqn. as,

$x {e}^{y} = \left(2 y - x\right) \left(y - 4\right) = 2 {y}^{2} - x y - 8 y + 4 x , \mathmr{and} ,$

$x {e}^{y} + x y + 8 y = 2 {y}^{2} + 4 x = 2 \left({y}^{2} + 2 x\right) .$

$\therefore \frac{d}{\mathrm{dx}} \left(x {e}^{y} + x y + 8 y\right) = \frac{d}{\mathrm{dx}} \left\{2 \left({y}^{2} + 2 x\right)\right\} = 2 \frac{d}{\mathrm{dx}} \left({y}^{2} + 2 x\right) .$

Following the Usual Rules of Diffn., we get,

$\therefore \frac{d}{\mathrm{dx}} \left(x {e}^{y}\right) + \frac{d}{\mathrm{dx}} \left(x y\right) + 8 \frac{d}{\mathrm{dx}} \left(y\right) = 2 \left\{\frac{d}{\mathrm{dx}} \left({y}^{2}\right) + 2 \frac{d}{\mathrm{dx}} \left(x\right)\right\} .$

Here, in the L.H.S., by the Product Rule, we have,

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

=xe^yd/dx(y)+e^y..........[because," the Chain Rule],"

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

Also, $\frac{d}{\mathrm{dx}} \left(x y\right) = x \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left(x\right) = x \frac{\mathrm{dy}}{\mathrm{dx}} + y .$

In the R.H.S., we have, $\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 2 y \frac{d}{\mathrm{dx}} \left(y\right) = 2 y \frac{\mathrm{dy}}{\mathrm{dx}} .$

Altogether, we get,

$x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} + x \frac{\mathrm{dy}}{\mathrm{dx}} + y + 8 \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left\{2 y \frac{\mathrm{dy}}{\mathrm{dx}} + 2\right\} = 4 y \frac{\mathrm{dy}}{\mathrm{dx}} + 4$

$\therefore \left(x {e}^{y} + x + 8 - 4 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 4 - {e}^{y} - y ,$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4 - {e}^{y} - y}{x {e}^{y} + x + 8 - 4 y} .$

Enloy Maths.!