3.5 Square Root and Cube Root

We've made good progress with our differential laws! Let's just look at two more: one for the square root and one for the cube root.

Law.

The differential law for square roots states

\[ \begin{align} \diff \bigl(\sqrt[2]{u}\bigr) &= \frac{1}{2 \mult \sqrt[2]{u}} \mult \diff u \end{align} \]

where \( u : \Number \) is a positive number, \( u \gt 0 \).

Why?

The squaring function and the square root function,

\[ \begin{align} b &\depends a & & & a &\depends b \\ b &= a^2 & &\andSpaced & a &= \sqrt[2]{b}, \end{align} \]

are inverse for positive numbers \( a : \Number \) and \( b : \Number \). We'll look for our differential law by rewriting the differential law for squaring.

\[ \begin{align} &\diff \bigl(a^2\bigr) = 2 a \mult \diff a \\ &\diff b = 2 a \mult \diff a \\ &\frac{1}{2 a} \mult \diff b = \diff a \\ &\frac{1}{2 \mult \sqrt[2]{b}} \mult \diff b = \diff \bigl(\sqrt[2]{b}\bigr) \end{align} \]

We've found it!

Law.

The differential law for cube roots states

\[ \begin{align} \diff \bigl(\sqrt[3]{u}\bigr) &= \frac{1}{3 \mult \bigl(\sqrt[3]{u}\bigr)^2} \mult \diff u \end{align} \]

where \( u : \Number \) is a non-zero number, \( u \neq 0 \).

Why?

The cubing function and the cube root function are inverse.

\[ \begin{align} b &\depends a & & & a &\depends b \\ b &= a^3 & &\andSpaced & a &= \sqrt[3]{b} \end{align} \]

Starting with the differential law for cubing, we'll try to find the differential law for the cube root.

\[ \begin{align} &\diff \bigl(a^3\bigr) = 3 a^2 \mult \diff a \\ &\diff b = 3 a^2 \mult \diff a \\ &\frac{1}{3 a^2} \mult \diff b = \diff a \\ &\frac{1}{3 \mult \bigl(\sqrt[3]{b}\bigr)^2} \mult \diff b = \diff \bigl(\sqrt[3]{b}\bigr) \end{align} \]

Got it.

Let's have one more example.

Example.

Consider the function

\[ \begin{align} z &\depends x \\ z &= \sqrt[2]{x^2 + 1}. \end{align} \]

Let's calculate the differential.

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