Partial Derivatives Examples 4

On the Partial Derivatives Examples 3 page, we looked at finding partial derivatives for more complicated functions. We will look at a few more examples of such functions.

Example 1

Consider the function $f(x, y, z) = \left\{\begin{matrix} \frac{xy^2z}{x^4 + y^4 + z^4} & \mathrm{if \: (x, y, z) \neq (0,0,0)}\\ 0 & \mathrm{if \: (x, y, z) = (0, 0, 0)} \end{matrix}\right.$. Evaluate $\frac{\partial}{\partial x} f(0, 0, 0)$, $\frac{\partial}{\partial y} f(0, 0, 0)$, and $\frac{\partial}{\partial z} f(0, 0, 0)$.

We will need to use the limit-definition of a partial derivative of a three variable function in order to obtain these partial derivatives. Recall that:

(1)

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

We first compute $\frac{\partial}{\partial x} f(0, 0, 0)$:

(2)

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

We also have that:

(3)

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

(4)

\begin{align} \quad \frac{\partial}{\partial z} f(0, 0, 0) = \lim_{h \to 0} \frac{f(0, 0, h) - f(0, 0, 0)}{h} = \lim_{h \to 0} \frac{0}{h} = 0 \end{align}

Example 1

Consider the function $f(x, y, z) = \left\{\begin{matrix} \frac{x + y^5 + z^2}{x^4 + y^4 + z^4} & \mathrm{if \: (x, y, z) \neq (0,0,0)}\\ 0 & \mathrm{if \: (x, y, z) = (0, 0, 0)} \end{matrix}\right.$. Evaluate $\frac{\partial}{\partial x} f(0, 0, 0)$, $\frac{\partial}{\partial y} f(0, 0, 0)$, and $\frac{\partial}{\partial z} f(0, 0, 0)$.

Using the limit-definition of a partial derivative and we have that:

(5)

\begin{align} \quad \frac{\partial}{\partial x} f(0, 0, 0) = \lim_{h \to 0} \frac{f(h, 0, 0) - f(0, 0, 0)}{h} = \lim_{h \to 0} \frac{\frac{h}{h^4}}{h} = \lim_{h \to 0} \frac{1}{h^4} = \infty \end{align}

(6)

\begin{align} \quad \frac{\partial}{\partial y} f(0, 0, 0) = \lim_{h \to 0} \frac{f(0, h, 0) - f(0, 0, 0)}{h} = \lim_{h \to 0} \frac{\frac{h^5}{h^4}}{h} = \lim_{h \to 0} 1 = 1 \end{align}

(7)

\begin{align} \quad \frac{\partial}{\partial z} f(0, 0, 0) = \lim_{h \to 0} \frac{f(0, 0, h) - f(0, 0,0)}{h} = \lim_{h \to 0} \frac{\frac{h^2}{h^4}}{h} = \lim_{h \to 0} \frac{1}{h^3} = \infty \end{align}

Note that the only first partial derivative that exists at the origin is $\frac{\partial}{\partial y} f(0, 0, 0)$.