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

##### 1 Answer
Sep 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(x - 5\right)}{2 \left(x + 1\right) \left(x - 2\right) \left(y - 1\right)}$.

#### Explanation:

Let us rewrite the given eqn. as, $: {e}^{y} \sqrt{x - 2} = x y + y = y \left(x + 1\right)$.

$\therefore \ln \left({e}^{y} \sqrt{x - 2}\right) = \ln \left\{y \left(x + 1\right)\right\}$.

$\therefore \ln {e}^{y} + \ln {\left(x - 2\right)}^{\frac{1}{2}} = \ln y + \ln \left(x + 1\right)$

$\therefore y + \frac{1}{2} \ln \left(x - 2\right) = \ln y + \ln \left(x + 1\right)$.

$\therefore \frac{d}{\mathrm{dx}} \left\{y + \frac{1}{2} \ln \left(x - 2\right)\right\} = \frac{d}{\mathrm{dx}} \left\{\ln y + \ln \left(x + 1\right)\right\}$.

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{2} \cdot \frac{1}{x - 2} = \frac{d}{\mathrm{dy}} \ln y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{1}{x + 1} \ldots \ldots \text{[Chain Rule]}$.

:. dy/dx+1/(2(x-2))=1/y*dy/dx+1/(x+1.

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

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

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(x - 5\right)}{2 \left(x + 1\right) \left(x - 2\right) \left(y - 1\right)}$.

Enjoy Maths.!