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3.3 Squaring and Cubing

Let's learn how to take the differential when squaring is involved!
Law.
The differential law for squaring states
\[ \begin{align} \diff \bigl(u^2\bigr) &= 2 u \mult \diff u \end{align} \]
where \( u : \Number \) is a number.
Why?
We'll check our law by rewriting the squaring function as a product: \( u^2 = u \mult u \).
\[ \begin{align} \diff \bigl(u^2\bigr) &= \diff (u \mult u) = u \mult \diff u + u \mult \diff u = 2 \mult \bigl(u \mult \diff u\bigr) = 2 u \mult \diff u \end{align} \]
We can now take the differential of any polynomial with degree two.
Example.
Consider the function
\[ \begin{align} z &\depends x \\ z &= 5 x^2 + 3 x - 1. \end{align} \]
Let's compute the differential.
\[ \begin{align} \diff z &= \diff \bigl(5 x^2 + 3 x - 1\bigr) \\ &= \diff \bigl(5 x^2 + 3 x\bigr) \\ &= \diff \bigl(5 x^2\bigr) + \diff (3 x) \\ &= 5 \mult \diff \bigl(x^2\bigr) + 3 \mult \diff x \\ &= 5 \mult 2 x \mult \diff x + 3 \mult \diff x \\ &= (10 x + 3) \mult \diff x \end{align} \]
We factored out the common differential term, \( \diff x \), in order to get our answer in standard form.
And now let's learn to take the differential of the cubing function.
Law.
The differential law for cubing states
\[ \begin{align} \diff \bigl(u^3\bigr) &= 3 u^2 \mult \diff u \end{align} \]
where \( u : \Number \) is a number.
Why?
We'll write \( u^3 \) as a product:
\[ \begin{align} u^3 &= u \mult u^2. \end{align} \]
We can then calculate.
\[ \begin{align} \diff \bigl(u^3\bigr) &= \diff \bigl(u \mult u^2\bigr) = u^2 \mult \diff u + u \mult \diff \bigl(u^2\bigr) = u^2 \mult \diff u + u \mult 2 u \mult \diff u = 3 u^2 \mult \diff u \end{align} \]
And again, let's have an example!
Example.
Let's take the differential of the function
\[ \begin{align} z &\depends x \\ z &= (2 x + 1)^3. \end{align} \]
We calculate.
\[ \begin{align} \diff z &= \diff \bigl((2 x + 1)^3\bigr) \\ &= 3 \mult (2 x + 1)^2 \mult \diff (2 x + 1) \\ &= 3 \mult (2 x + 1)^2 \mult \diff (2 x) \\ &= 6 \mult (2 x + 1)^2 \mult \diff x \end{align} \]
Notice how we matched
\[ \begin{align} u &= 2 x + 1 \end{align} \]
when we applied the differential law for cubing.
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