Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=\arctan \frac{y}{x} $$

Short answer

The four second partial derivatives are \(\frac{\partial^2 z}{\partial x^2} = \frac{2x^2y-y^3}{{(x^2+y^2)}^3}\), \(\frac{\partial^2 z}{\partial y^2} = \frac{y^3-2x^2y}{{(x^2+y^2)}^3}\), and the second mixed partials \(\frac{\partial^2 z}{\partial x\partial y} = \frac{\partial^2 z}{\partial y\partial x} = \frac{-2xy}{{(x^2+y^2)}^2}\)

Step-by-step solution

Calculate the First Partial Derivative with Respect to x

The first partial derivative of \(z\) with respect to \(x\), denoted as \(\frac{\partial z}{\partial x}\), is calculated using the quotient rule, which yields: \[ \frac{\partial z}{\partial x} =\frac{-y}{{x^2}+{y^2}} \]

Calculate the First Partial Derivative with Respect to y

The first partial derivative of \(z\) with respect to \(y\), denoted as \(\frac{\partial z}{\partial y}\), is calculated using the quotient rule, which leads to: \[ \frac{\partial z}{\partial y} =\frac{x}{{x^2}+{y^2}} \]

Calculate the Second Partial Derivatives

Now, from the first order partial derivatives, the second order partial derivatives with respect to x and y are calculated as: \[ \frac{\partial^2 z}{\partial x^2} \text{ and } \frac{\partial^2 z}{\partial y^2} \] After calculating, we get \[ \frac{\partial^2 z}{\partial x^2} =\frac{2x^2y-y^3}{{(x^2+y^2)}^3} \] and \[ \frac{\partial^2 z}{\partial y^2} =\frac{y^3-2x^2y}{{(x^2+y^2)}^3} \]

Observe that the Second Mixed Partials are equal

The two mixed second partial derivatives i.e., \(\frac{\partial^2 z}{\partial x\partial y}\) and \(\frac{\partial^2 z}{\partial y\partial x}\) can be found from our first partial derivatives. After calculating, we observe that both equal: \[ \frac{\partial^2 z}{\partial x\partial y} =\frac{-2xy}{{(x^2+y^2)}^2} = \frac{\partial^2 z}{\partial y\partial x} \] which verifies Clairaut's theorem of mixed partials being equal.