## Calculus 8th Edition

$y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$
Given: $z=y+f(x^2-y^2)$ Need to prove that $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$ Now, $\dfrac{\partial z}{\partial x}=2xf'(x^2-y^2)$ and $\dfrac{\partial z}{\partial y}=1-2yf'(x^2-y^2)$ Thus, $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$ or, $y(2xf'(x^2-y^2))+x(1-2yf'(x^2-y^2))=x$ or $2xyf'(x^2-y^2)+x-2xyf'(x^2-y^2)+x=x$ or, $x=x$ Hence, it has been proved that $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=x$