If z = y + f(v),where v = frac{x}{y},then v f | vedclass

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(B) Given $z = y + f(v)$ and $v = \frac{x}{y}$.Substituting $v$,we have $z = y + f\left(\frac{x}{y}\right)$.Now,differentiate $z$ partially with respe

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$1$

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(B) Given $z = y + f(v)$ and $v = \frac{x}{y}$.Substituting $v$,we have $z = y + f\left(\frac{x}{y}\right)$.Now,differentiate $z$ partially with respect to $x$:$\frac{\partial z}{\partial x} = f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}$.Next,differentiate $z$ partially with respect to $y$ using the chain rule:$\frac{\partial z}{\partial y} = 1 + f'\left(\frac{x}{y}\right) \cdot \left(-\frac{x}{y^2}\right) = 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.Now,calculate the expression $v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$:$v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = \left(\frac{x}{y}\right) \left[f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}\right] + 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.$= \frac{x}{y^2} f'\left(\frac{x}{y}\right) + 1 - \frac{x}{y^2} f'\left(\frac{x}{y}\right)$.$= 1$.

If $z = y + f(v)$,where $v = \frac{x}{y}$,then $v \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}$ is

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