If u=e^{x^2-y^2},then | vedclass

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

Question

(C) Given that,$u=e^{x^2-y^2}$.First,find the partial derivative of $u$ with respect to $x$:$u_x = \frac{\partial}{\partial x}(e^{x^2-y^2}) = e^{x^2-y

Vedclass · Accepted Answer

$y u_x+x u_y=0$

Vedclass · Answer

(C) Given that,$u=e^{x^2-y^2}$.First,find the partial derivative of $u$ with respect to $x$:$u_x = \frac{\partial}{\partial x}(e^{x^2-y^2}) = e^{x^2-y^2}(2x)$.Next,find the partial derivative of $u$ with respect to $y$:$u_y = \frac{\partial}{\partial y}(e^{x^2-y^2}) = e^{x^2-y^2}(-2y)$.Now,multiply $u_x$ by $y$:$y u_x = y \cdot e^{x^2-y^2}(2x) = 2xy e^{x^2-y^2}$.Multiply $u_y$ by $x$:$x u_y = x \cdot e^{x^2-y^2}(-2y) = -2xy e^{x^2-y^2}$.Adding these two expressions:$y u_x + x u_y = 2xy e^{x^2-y^2} - 2xy e^{x^2-y^2} = 0$.Thus,the correct option is $C$.

If $u=e^{x^2-y^2}$,then

Similar Questions

If $f(x, y) = \frac{\cos(x - 4y)}{\cos(x + 4y)}$,then $\left. \frac{\partial f}{\partial x} \right|_{y = \frac{x}{2}}$ is equal to:

If $z = \sec^{-1}\left(\frac{x^4+y^4-8x^2y^2}{x^2+y^2}\right)$, then $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y}$ is equal to

$\begin{aligned} & f(x, y)=2(x-y)^2-x^4-y^4 \\ & \left|\left(f_{x x} f_{y y}-f_{x y}^2\right)\right|_{(0,0)} \end{aligned}$

If $u = \log (x^3 + y^3 + z^3 - 3xyz)$,then $\left( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} \right) (x + y + z) =$

If $z=\log (\tan x+\tan y)$,then $(\sin 2 x) \frac{\partial z}{\partial x}+(\sin 2 y) \frac{\partial z}{\partial y}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module