If u = frac{x + y}{x - y},then frac{partial u | vedclass

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

Question

(B) Given $u = \frac{x + y}{x - y}$.First,find the partial derivative of $u$ with respect to $x$ using the quotient rule:$\frac{\partial u}{\partial x

Vedclass · Accepted Answer

$\frac{2}{x - y}$

Vedclass · Answer

(B) Given $u = \frac{x + y}{x - y}$.First,find the partial derivative of $u$ with respect to $x$ using the quotient rule:$\frac{\partial u}{\partial x} = \frac{(x - y)(1) - (x + y)(1)}{(x - y)^2} = \frac{x - y - x - y}{(x - y)^2} = \frac{-2y}{(x - y)^2}$.Next,find the partial derivative of $u$ with respect to $y$ using the quotient rule:$\frac{\partial u}{\partial y} = \frac{(x - y)(1) - (x + y)(-1)}{(x - y)^2} = \frac{x - y + x + y}{(x - y)^2} = \frac{2x}{(x - y)^2}$.Now,add the two partial derivatives:$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = \frac{-2y}{(x - y)^2} + \frac{2x}{(x - y)^2} = \frac{2x - 2y}{(x - y)^2} = \frac{2(x - y)}{(x - y)^2} = \frac{2}{x - y}$.

If $u = \frac{x + y}{x - y}$,then $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = $

Similar Questions

If $u=\log \left(x^3+y^3+z^3-3 x y z\right)$,then $(x+y+z)(u_x+u_y+u_z)$ is equal to

If $z = y + f(v)$,where $v = \frac{x}{y}$,then $v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$ is

If $u = \tan^{-1}(x + y),$ then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = $

If $u = \log_e(x^2 + y^2) + \tan^{-1}\left(\frac{y}{x}\right)$,then $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = $

If $u(x, y)=y \log x+x \log y$,then $u_x u_y-u_x \log x-u_y \log y+\log x \log y$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module