The equation y^2e^{xy} = 9e^{-3}x^2 defines y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $y^2e^{xy} = 9e^{-3}x^2$. Differentiating both sides with respect to $x$ using the product rule and chain rule: $\frac{d}{dx}(

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) Given the equation: $y^2e^{xy} = 9e^{-3}x^2$. Differentiating both sides with respect to $x$ using the product rule and chain rule: $\frac{d}{dx}(y^2)e^{xy} + y^2\frac{d}{dx}(e^{xy}) = 9e^{-3}\frac{d}{dx}(x^2)$ $2y\frac{dy}{dx}e^{xy} + y^2e^{xy}\left(y + x\frac{dy}{dx}ight) = 18e^{-3}x$ Substitute $x = -1$ and $y = 3$: $2(3)\frac{dy}{dx}e^{-3} + (3)^2e^{-3}\left(3 + (-1)\frac{dy}{dx}ight) = 18e^{-3}(-1)$ Divide the entire equation by $e^{-3}$ (since $e^{-3} 
eq 0$): $6\frac{dy}{dx} + 9(3 - \frac{dy}{dx}) = -18$ $6\frac{dy}{dx} + 27 - 9\frac{dy}{dx} = -18$ $-3\frac{dy}{dx} = -18 - 27$ $-3\frac{dy}{dx} = -45$ $\frac{dy}{dx} = 15$.

The equation $y^2e^{xy} = 9e^{-3}x^2$ defines $y$ as a differentiable function of $x$. The value of $\frac{dy}{dx}$ for $x = -1$ and $y = 3$ is

Similar Questions

If $3\sin (xy) + 4\cos (xy) = 5$,then $\frac{dy}{dx} = $

Find $\frac{dy}{dx}$,if $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}$.

If $(x^2-3x+2) e^{\frac{y}{x-1}}=x+2$, then find the value of $(\frac{dy}{dx})_{x=0}$.

If $\log (x+y)=\log (x y)+a$,where $a$ is a constant,then $\frac{dy}{dx}$ at $x=2$ and $y=4$ is

If $x \sqrt{1+y}+y \sqrt{1+x}=0$,then $\frac{d y}{d x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module