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

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(B) Given $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$. First,calculate $\frac{\partial z}{\partial x}$: $\

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

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(B) Given $z = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$. First,calculate $\frac{\partial z}{\partial x}$: $\frac{\partial z}{\partial x} = \frac{y}{x} \left[ \cos \frac{x}{y} \cdot \frac{1}{y} - \sin \left( 1 + \frac{y}{x} \right) \cdot \left( -\frac{y}{x^2} \right) \right] - \frac{y}{x^2} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$ Multiplying by $x$: $x \frac{\partial z}{\partial x} = \cos \frac{x}{y} + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$ $x \frac{\partial z}{\partial x} = \cos \frac{x}{y} + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - z$ ... $(i)$ Now,calculate $\frac{\partial z}{\partial y}$: $\frac{\partial z}{\partial y} = \frac{1}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] + \frac{y}{x} \left[ \cos \frac{x}{y} \cdot \left( -\frac{x}{y^2} \right) - \sin \left( 1 + \frac{y}{x} \right) \cdot \frac{1}{x} \right]$ Multiplying by $y$: $y \frac{\partial z}{\partial y} = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] - \cos \frac{x}{y} - \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right)$ $y \frac{\partial z}{\partial y} = z - \cos \frac{x}{y} - \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right)$ ... $(ii)$ Adding $(i)$ and $(ii)$: $x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 0$ Therefore,$x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

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

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