Calculate \(\frac{\partial z}{\partial x}\) when (a) \(z=\frac{y}{x^{2}}-\frac{x}{y^{2}}\), (b) \(z=\mathrm{e}^{x^{2}-4 x y}\)

To find the partial derivative, we treat y as a constant and differentiate z with respect to x: 1) Differentiate the first term \(\frac{y}{x^{2}}\) with respect to x: \(\frac{\partial}{\partial x}\left(\frac{y}{x^{2}}\right) = y \cdot \frac{\partial}{\partial x}\left(\frac{1}{x^{2}}\right)\) Using the power rule for differentiation: \(\frac{\partial}{\partial x}\left(\frac{1}{x^{2}}\right) = -2x^{-3}\) So, the derivative of the first term is: \(y \cdot -2x^{-3} = -\frac{2y}{x^{3}}\) 2) Differentiate the second term \(\frac{x}{y^{2}}\) with respect to x: \(\frac{\partial}{\partial x}\left(\frac{x}{y^{2}}\right) = \frac{1}{y^{2}} \cdot \frac{\partial}{\partial x}(x)\) Since the derivative of x with respect to x is 1: \(\frac{\partial}{\partial x}(x) = 1\) So, the derivative of the second term is: \(\frac{1}{y^{2}} \cdot 1 = \frac{1}{y^{2}}\) 3) Combine the derivatives of the two terms, and we get: \(\frac{\partial z}{\partial x} = -\frac{2y}{x^{3}} - \frac{1}{y^{2}}\)

To find the partial derivative, we treat y as a constant and differentiate z with respect to x using the chain rule: 1) Differentiate \(e^{x^{2}-4xy}\) with respect to x: \(\frac{\partial}{\partial x}\left(e^{x^{2}-4xy}\right) = e^{x^{2}-4xy} \cdot \frac{\partial}{\partial x}\left(x^{2}-4xy\right)\) 2) Differentiate the exponent in the expression, treating y as a constant: \(\frac{\partial}{\partial x}\left(x^{2}-4xy\right) = 2x - 4y\) 3) Plug the derivative of the exponent back into the original expression: \(\frac{\partial z}{\partial x} = e^{x^{2}-4xy} \cdot (2x - 4y)\)