Question

If $x^2.e^y+2x{ye}^x+13=0$, then $\frac{dy}{dx}$ is

• A. $\frac{-2x{e}^{y-x}-2y(x+1)}{x({xe}^{y-x}+2)}$
• B. $\frac{2x{e}^{x-y}+2y(x+1)}{x({xe}^{y-x}+2)}$
• C. $\frac{-2x{e}^{x-y}+2y(x+1)}{x({xe}^{y-x}+2)}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{-2x{e}^{y-x}-2y(x+1)}{x({xe}^{y-x}+2)}$

Differentiating w.r.t. $x$

$2x{e}^y+x^2{e}^y\frac{dy}{dx}+2\left[\frac{d}{dx}\right(xy)e^x+xy{e}^x]=0$

$2x{e}^y+x^2{e}^y\frac{dy}{dx}+2\left[\frac{d}{dx}\right(xy)e^x+xy{e}^x]=0$

$2x{e}^y+x^2{e}^y\frac{dy}{dx}+2e^x[y+x\frac{dy}{dx}+xy]=0$

$2x{e}^y+x^2{e}^y\frac{dy}{dx}+2y{e}^x+2x{e}^x\frac{dy}{dx}+2xy2e^x=0$

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

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

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

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