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

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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} = $

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