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

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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

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