# Partial Derivatives Examples 3

We have just looked at some examples of determining partial derivatives of a function from the Partial Derivatives Examples 1 and Partial Derivatives Examples 2 page. We will now look at finding partial derivatives for more complex functions.

## Example 1

**Determine the first partial derivatives of the function $f(x, y) = \left\{\begin{matrix}\frac{2x^3-y^3}{x^2 + 3y^2} & \mathrm{if} \: (x, y) \neq (0, 0))\\ 0 & \mathrm{if} (x, y) = (0, 0) \end{matrix}\right.$, and determine if these partial derivatives are continuous at $(0, 0)$.**

Notice that in this example, $f$ is defined for all $(x, y) \in \mathbb{R}^2$, however, this function is a piecewise function, so we will deal with the partial derivatives of each piece of $f$ individually.

First suppose that $(x, y) \neq (0, 0)$. Then we can compute the partial derivatives of $f$ normally by applying the quotient rule:

(1)Now suppose that $(x, y) = (0, 0)$. We cannot use the partial derivative formulas above because we would get division by $0$, so instead, we will use the formal definition of the partial derivative at $(0, 0)$.

(2)We will now determine whether or not these partial derivatives are continuous at $(0, 0) \in \mathbb{R}^2$. To show that, we must show whether or not $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial x} (x, y) = 2$ and whether or not $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial y} (x, y) = -\frac{1}{3}$.

First let's determine if $\frac{\partial f}{\partial x}$ is continuous at $(0, 0)$.

(4)Now notice that if we evaluate this limit along the line $x = 0$ then we get that $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial x} (x, y) = 0$ (which already tells us that our partial derivative is discontinuous), but also, if we evaluate this limit along the line $y = 0$ then we get that $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial x} (x, y) = 2$. Since we have two different limits, we conclude that $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial x} (x, y)$ does not exist. Therefore we have that $\frac{\partial f}{\partial y}$ is discontinuous at $(0, 0)$.

Now let's determine if $\frac{\partial f}{\partial y}$ is continuous at $(0, 0)$.

(5)Now notice that if we evaluate this limit along the line $x = 0$ we get that $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial y} (x, y) = - \frac{1}{3}$. However, if we evaluate this limit along the line $y = 0$ we get that $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial y} (x, y) = 0$. Since we have two different limits, we conclude that $\lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial y} (x, y)$ does not exist. Therefore we have that $\frac{\partial f}{\partial y}$ is discontinuous at $(0, 0)$.

*Note that from Example 1 above, that a function can be differentiable even though its partial derivatives may be discontinuous.*

## Example 2

Determine the first partial derivatives of the function $f(x, y) = \left\{\begin{matrix}\frac{(x+y)^3}{x^2 + y^2} & \mathrm{if} \: (x, y) \neq (0, 0))\\ 0 & \mathrm{if} (x, y) = (0, 0) \end{matrix}\right.$.**

First suppose that $(x, y) \neq 0$. Then the partial derivatives of $f$ are:

(6)Now if $(x, y) \neq 0$, then:

(7)