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)

\begin{align} \quad \frac{\partial f}{\partial x} = \frac{(x^2 + 3y^2)(6x^2) - (2x^3 - y^3)(2x)}{(x^2 + 3y^2)^2} \quad , \quad \frac{\partial f}{\partial y} = \frac{(x^2 + 3y^2)(-3y^2) - (2x^3 - y^3)(6y)}{(x^2 + 3y^2)^2} \end{align}

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)

\begin{align} \quad \frac{\partial f}{\partial x} (0, 0) = \lim_{h \to 0} \frac{f(0 + h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \frac{\frac{2h^3}{h} - 0}{h} = \lim_{h \to 0} \frac{2h^3}{h^3} = 2 \end{align}

(3)

\begin{align} \quad \frac{\partial f}{\partial y} (0, 0) = \lim_{h \to 0} \frac{f(0, 0 + h) - f(0, 0)}{h} = \lim_{h \to 0} \frac{-\frac{h^3}{3h^2} - 0}{h} = \lim_{h \to 0} \frac{-h^3}{3h^3} = -\frac{1}{3} \end{align}

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)

\begin{align} \quad \lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial x} (x, y) = \lim_{(x, y) \to (0, 0)} \frac{(x^2 + 3y^2)(6x^2) - (2x^3 - y^3)(2x)}{(x^2 + 3y^2)^2} = \lim_{(x, y) \to (0, 0)} \frac{2x^4 + 18x^2y^2 + 2xy^3}{x^4 + 6x^2y^2 + 9y^4} \end{align}

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)

\begin{align} \quad \lim_{(x, y) \to (0, 0)} \frac{\partial f}{\partial y} (x, y) = \lim_{(x, y) \to (0, 0)} \frac{(x^2 + 3y^2)(-3y^2) - (2x^3 - y^3)(6y)}{(x^2 + 3y^2)^2} = \lim_{(x, y) \to (0, 0)} \frac{-3x^2y^2 - 3y^4 - 12x^3y}{x^4 + 6x^2y^2 + 9y^4} \end{align}

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)

\begin{align} \quad \frac{\partial f}{\partial x} = \frac{3(x^2 + y^2)(x + y)^2 - (x+y)^3(2x)}{(x^2 + y^2)^2} \quad , \quad \frac{\partial f}{\partial y} = \frac{3(x^2 + y^2)(x + y)^2 - (x + y)^3(2y)}{(x^2 + y^2)^2} \end{align}

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

(7)

\begin{align} \quad \frac{\partial f}{\partial x} (0, 0) = \lim_{h \to 0} \frac{f(0 + h, 0) - f(0, 0)}{h} = \lim_{h \to 0} \frac{h^3}{h^3} = 1 \end{align}

(8)

\begin{align} \quad \frac{\partial f}{\partial y} (0, 0) = \lim_{h \to 0} \frac{f(0, 0+h) - f(0, 0)}{h} = \lim_{h \to 0} \frac{h^3}{h^3} = 1 \end{align}