91Ó°ÊÓ

Given the function \[ f(x, y) = \frac{x}{y^2} - \frac{y}{x^2} \]we need to calculate the partial derivative of \(f\) with respect to \(x\). Use the quotient rule for differentiation: \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \]For the first term, \(u = x\) and \(v = y^2\), yielding: \[ \frac{\partial}{\partial x} \left( \frac{x}{y^2} \right) = \frac{y^2 \cdot 1 - x \cdot 0}{y^4} = \frac{1}{y^2} \]For the second term, \(u = y\) and \(v = x^2\), yielding:\[ \frac{\partial}{\partial x} \left( -\frac{y}{x^2} \right) = -\left( \frac{x^2 \cdot 0 - y \cdot 2x}{x^4} \right) = \frac{2y}{x^3} \]Thus, the first partial derivative \(f_x\) is:\[ f_x = \frac{1}{y^2} + \frac{2y}{x^3} \]

Next, we find the partial derivative of \(f\) with respect to \(y\). Applying the quotient rule again to each term:For the first term, \(u = x\) and \(v = y^2\), we obtain:\[ \frac{\partial}{\partial y} \left( \frac{x}{y^2} \right) = \frac{y^2 \cdot 0 - x \cdot 2y}{y^4} = -\frac{2xy}{y^{4}} = -\frac{2x}{y^3} \]For the second term, \(u = y\) and \(v = x^2\), we get:\[ \frac{\partial}{\partial y} \left( -\frac{y}{x^2} \right) = -\left( \frac{x^2 \cdot 1}{x^4} \right) = -\frac{1}{x^2} \]Thus, the first partial derivative \(f_y\) is:\[ f_y = -\frac{2x}{y^3} - \frac{1}{x^2} \]

First, find \(f_{xy}\) by differentiating \(f_x\) with respect to \(y\):For \(f_x = \frac{1}{y^2} + \frac{2y}{x^3}\), we get:\[ \frac{\partial}{\partial y} \left( \frac{1}{y^2} \right) = -\frac{2}{y^3} \]For the second term, differentiate \(\frac{2y}{x^3}\) with respect to \(y\):\[ \frac{\partial}{\partial y} \left( \frac{2y}{x^3} \right) = \frac{2}{x^3} \]Thus, \[ f_{xy} = -\frac{2}{y^3} + \frac{2}{x^3} \]

Next, find \(f_{yx}\) by differentiating \(f_y\) with respect to \(x\):For \( f_y = -\frac{2x}{y^3} - \frac{1}{x^2} \), differentiate the first term:\[ \frac{\partial}{\partial x} \left( -\frac{2x}{y^3} \right) = -\frac{2}{y^3} \]For the second term, \[ \frac{\partial}{\partial x} \left( -\frac{1}{x^2} \right) = \frac{2}{x^3} \]Thus,\[ f_{yx} = -\frac{2}{y^3} + \frac{2}{x^3} \]

For the second derivative \(f_{xx}\), differentiate \(f_x\) with respect to \(x\):For the first term \(\frac{1}{y^2}\), the derivative is 0 since it is a constant with respect to \(x\).For \(\frac{2y}{x^3}\), apply the power rule:\[ \frac{\partial}{\partial x} \left( \frac{2y}{x^3} \right) = \frac{2y \cdot (-3)}{x^4} = -\frac{6y}{x^4} \]Thus, \[ f_{xx} = -\frac{6y}{x^4} \]

Finally, for \(f_{yy}\), differentiate \(f_y\) with respect to \(y\):For \(-\frac{2x}{y^3}\), apply the power rule:\[ \frac{\partial}{\partial y} \left( -\frac{2x}{y^3} \right) = \frac{6x}{y^4} \]The \(\frac{1}{x^2}\) term is a constant with respect to \(y\), so its derivative is 0.Thus, \[ f_{yy} = \frac{6x}{y^4} \]