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 $\frac{\partial z}{\partial x}$:$\f

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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 \left( \frac{x}{y} \right) \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]$$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right]$$x \frac{\partial z}{\partial x} = \cos \left( \frac{x}{y} \right) + \frac{y^2}{x^2} \sin \left( 1 + \frac{y}{x} \right) - z \quad \dots (i)$Next, 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 \left( \frac{x}{y} \right) \cdot \left( -\frac{x}{y^2} \right) - \sin \left( 1 + \frac{y}{x} \right) \cdot \frac{1}{x} \right]$$y \frac{\partial z}{\partial y} = \frac{y}{x} \left[ \sin \frac{x}{y} + \cos \left( 1 + \frac{y}{x} \right) \right] - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right)$$y \frac{\partial z}{\partial y} = z - \cos \left( \frac{x}{y} \right) - \frac{y}{x} \sin \left( 1 + \frac{y}{x} \right) \quad \dots (ii)$Comparing $(i)$ and $(ii)$, we observe that $x \frac{\partial z}{\partial x} = -y \frac{\partial z}{\partial y}$.

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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