If z = frac{y}{x} left[ sin left( frac{x}{y} | vedclass

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(B) The given function is $z = f\left( \frac{x}{y} \right)$. This implies that $z$ is a homogeneous function of $x$ and $y$ with degree $n = 0$. Accor

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$-y \frac{\partial z}{\partial y}$

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(B) The given function is $z = f\left( \frac{x}{y} \right)$. This implies that $z$ is a homogeneous function of $x$ and $y$ with degree $n = 0$. According to Euler's theorem for homogeneous functions,if $z$ is a homogeneous function of degree $n$,then $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = nz$. Since $n = 0$,we have $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0 \cdot z = 0$. Therefore,$x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

If $z = \frac{y}{x} \left[ \sin \left( \frac{x}{y} \right) + \cos \left( 1 + \frac{y}{x} \right) \right]$,then $x \frac{\partial z}{\partial x} = $

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