5.2 The Second Derivative

When we take the derivative of a function \( z \depends x \), the result, \( \overset{x}{\diff} z \depends x \), is also a function. We find the second derivative of \( z \) by taking another derivative.

\[ \begin{align} \overset{x}{\diff} \overset{x}{\diff} z &\depends x \end{align} \]

And we could keep going, taking the derivative again and again!

Example.

Let's compute the first two derivatives of the function

\[ \begin{align} z &\depends x \\ z &= x^3 + 5 x^2. \end{align} \]

Here's the first \( x \)-derivative.

\[ \begin{align} \overset{x}{\diff} z &= \overset{x}{\diff} \bigl(x^3 + 5 x^2\bigr) \\ &= \overset{x}{\diff} \bigl(x^3\bigr) + \overset{x}{\diff}\bigl(5 x^2\bigr) \\ &= 3 x^2 + 5 \mult \overset{x}{\diff} \bigl(x^2\bigr) \\ &= 3 x^2 + 5 \mult 2 x \mult \overset{x}{\diff} x \\ &= 3 x^2 + 10 x \end{align} \]

Now let's take the \( x \)-derivative, again.

\[ \begin{align} \overset{x}{\diff} \overset{x}{\diff} z &= \overset{x}{\diff} \bigl(3 x^2 + 10 x\bigr) \\ &= \overset{x}{\diff} \bigl(3 x^2\bigr) + \overset{x}{\diff} \bigl(10 x\bigr) \\ &= 3 \mult \overset{x}{\diff} \bigl(x^2\bigr) + 10 \mult \overset{x}{\diff} x \\ &= 3 \mult 2 x \mult \overset{x}{\diff} x + 10 \\ &= 6 x + 10 \end{align} \]

We've found our function's second derivative!

Let's take the second derivative of the square root function.

Example.

The square root function,

\[ \begin{align} z &\depends x \\ z &= \sqrt[2]{x}, \end{align} \]

is defined for positive inputs, \( x \ge 0 \). We've seen that the square root has first derivative

\[ \begin{align} \overset{x}{\diff} z &= \frac{1}{2 \mult \sqrt[2]{x}}. \end{align} \]

Let's calculate the second derivative.

\[ \begin{align} \overset{x}{\diff} \overset{x}{\diff} z &= \overset{x}{\diff} \biggl(\frac{1}{2 \mult \sqrt[2]{x}}\biggr) \\ &= \frac{1}{2} \mult \overset{x}{\diff} \biggl(\frac{1}{\sqrt[2]{x}}\biggr) \\ &= \frac{1}{2} \mult \frac{-1}{\bigl(\sqrt[2]{x}\bigr)^2} \mult \overset{x}{\diff} \Bigl(\sqrt[2]{x}\Bigr) \\ &= \frac{1}{2} \mult \frac{-1}{\bigl(\sqrt[2]{x}\bigr)^2} \mult \frac{1}{2 \mult \sqrt[2]{x}} \mult \overset{x}{\diff} x \\ &= \frac{-1}{4 x \mult \sqrt[2]{x}} \end{align} \]

Like the first derivative, the second derivative is defined only for \( x \gt 0 \).

For an easy exercise, you can calculate the second derivative of the cubing function.

\[ \begin{align} \overset{x}{\diff} \overset{x}{\diff} \bigl(x^3\bigr) &= 6 x \end{align} \]

For a much more challenging exercise, calculate the second derivative of the cube root function.

\[ \begin{align} \overset{x}{\diff} \overset{x}{\diff} \Bigl(\sqrt[3]{x}\Bigr) &= \frac{-2}{9 \mult \bigl(\sqrt[3]{x}\bigr)^5} \end{align} \]

We'll find an easier way to calculate this cube root exercise when we study base functions.