Question

If $x^2{e}^y+2xy{e}^x+23=0$ then $\frac{dy}{dx}=$

• A. $2x{e}^{y-x}+2y(x+1)$
• B. $2x{e}^{x-y}-3y(x+1)$
• C. $\frac{2x{e}^{y-x}-2y(x+1)}{x(x{e}^{y-x}+2)}$
• D. $2x{e}^{y-x}-y(x+1)$

Answer 1

The correct option is C

$\frac{2x{e}^{y-x}-2y(x+1)}{x(x{e}^{y-x}+2)}$

Explanation for the correct option

Let $y=x^2{e}^y+2xy{e}^x+23=0$, then

$$\begin{gathered} 2x{e}^y+x^2{e}^y\frac{dy}{dx}+2\left[\frac{d\left(xy{e}^x\right)}{dx}+xy{e}^x\right]=0 \\ \Rightarrow 2x{e}^y+x^2{e}^y\frac{dy}{dx}+2e^x\left[y+x\frac{dy}{dx}+xy\right]=0 \\ \Rightarrow 2x{e}^y+x^2{e}^y\frac{dy}{dx}+2y{e}^x+2x{e}^x\frac{dy}{dx}+2xy2e^x=0 \\ \Rightarrow \frac{dy}{dx}=\frac{2x{e}^{y-x}-2y\left(1+x\right)}{x\left(x{e}^{y-x}+2\right)} \end{gathered}$$

Hence, option C is correct.