Section2.6Using the Arithmetic of Derivatives – Examples
Subsection2.6.1Using the Arithmetic of Derivatives – Examples
In this section we illustrate the computation of derivatives using the arithmetic of derivatives — Theorems 2.4.2, 2.4.3 and 2.4.5. To make it clear which rules we are using during the examples we will note which theorem we are using:
If we can find the derivatives of \(f_1(x)\) and \(f_2(x)\text{,}\) we will be able to get the derivative of \(f(x)\) just by applying the quotient rule. The derivative, \(f'_1(x)=1\text{,}\) of \(f_1(x)\) is easy, so let’s work on \(f_2(x)\text{.}\)
If we can find the derivatives of \(f_3(x)\) and \(f_4(x)\text{,}\) we will be able to get the derivative of \(f_2(x)\) just by applying linearity (Theorem 2.4.2). The derivative, \(f'_3(x)=1\text{,}\) of \(f_3(x)\) is easy. So let’s work of \(f_4(x)\text{.}\)
If we can find the derivative of \(f_5(x)\text{,}\) we will be able to get the derivative of \(f_4(x)\) just by applying the special case the quotient rule (Corollary 2.4.6). The derivative of \(f_5(x)\) is easy.
So we have completed breaking down \(f(x)\) into easy pieces. It is now just a matter of reversing the break down steps, putting everything back together, starting with the easy pieces and working up to \(f(x)\text{.}\) Here goes.
We now have an answer. But we really should clean it up, not only to make it easier to read, but also because invariably such computations are just small steps inside much larger computations. Any future computations involving this expression will be a lot easier and less error prone if we clean it up now. Cancelling the \(2x\) and the \(-2x\) in
While the linearity theorem (Theorem 2.4.2) is stated for a linear combination of two functions, it is not difficult to extend it to linear combinations of three or more functions as the following example shows.
Example2.6.5.Linearity of the derivative of three or more functions.
We’ll start by generalising linearity to three functions.
\begin{align*}
\diff{}{x}\big\{a F(x)+ b G(x)+ c H(x)\big\}
&=\diff{}{x}\big\{\ a \cdot [F(x)]\ +\ 1\cdot[b G(x)+c H(x)]\ \big\}\\
&=a F'(x)+\diff{}{x}\{b G(x)+c H(x)\}\\
&\qquad\text{by LIN with $\alpha=a, f(x)=F(x), \beta=1$,}\\
&\qquad\text{and } g(x)=b G(x)+c H(x)\\
&=a F'(x)+b G'(x)+c H'(x)\\
&\qquad\text{by LIN with } \alpha=b, f(x)=G(x), \beta=c,\\
&\qquad\text{and } g(x)=H(x)
\end{align*}
This gives us linearity for three terms, namely (just replacing upper case names by lower case names)
Similarly, while the product rule is stated for the product of two functions, it is not difficult to extend it to the product of three or more functions as the following example shows.
Example2.6.6.Extending the product rule to more than two factors.
Once again, we’ll start by generalising the product rule to three factors.
Observe that when we differentiate a product of three factors, the answer is a sum of three terms and in each term the derivative acts on exactly one of the original factors. Just by repeating the above argument many times, we may generalise the product rule to give the derivative of a product of \(n\) factors, for any natural number \(n\text{:}\)
When we differentiate a product of \(n\) factors, the answer is a sum of \(n\) terms and in each term the derivative acts on exactly one of the original factors. In the first term, the derivative acts on the first of the original factors. In the second term, the derivative acts on the second of the original factors. And so on.
If we make \(f_1(x) = f_2(x) = \cdots = f_n(x) = f(x)\) then each of the \(n\) terms on the right hand side of the above equation is the product of \(f'(x)\) and exactly \(n-1\)\(f(x)\)’s, and so is exactly \(f(x)^{n-1}\,f'(x)\text{.}\) So we get the following useful result
We can now compute the derivative of \(x^n\) for any natural number \(n\text{.}\) Start with Lemma 2.6.7 and substitute \(f(x)=x\) and \(f'(x)=1\text{:}\)
Again — this is a result we will come back to quite a few times in the future, so we should make sure we can refer to it easily. However, at present this statement only holds when \(n\) is a positive integer. With a little more work we can extend this to compute \(x^q\) where \(q\) is any positive rational number and then any rational number at all (positive or negative). So let us hold off for a little longer. Instead we can make it a lemma, since it will be an ingredient in quite a few of the examples following below and in constructing the final corollary.
Lemma2.6.9.Derivative of \(x^n\).
Let \(n\) be a positive integer then
\begin{equation*}
\diff{}{x} x^n = n x^{n-1}
\end{equation*}
Back to more examples.
Example2.6.10.Derivative of a polynomial.
\begin{align*}
\diff{}{x}\big\{2x^3+4x^5\big\}
&=2\diff{}{x}\{x^3\}+4\diff{}{x}\{x^5\}\\
&\qquad\qquad\text{by LIN with } \alpha=2, f(x)=x^3, \beta=4 \text{ and } g(x)=x^5\\
&=2\{3x^2\}+4\{5x^4\}\\
\end{align*}
by Lemma 2.6.9, once with \(n=3\text{,}\) and once with \(n=5\)
\begin{align*}
&=6x^2+20x^4
\end{align*}
Example2.6.11.Derivative of product of polynomials.
In this example we’ll compute \(\diff{}{x}\big\{(3x+9)(x^2+4x^3)\big\}\) in two different ways. For the first, we’ll start with the product rule.
\begin{align*}
\diff{}{x}\bigg\{\frac{4x^3-7x}{4x^2+1}\bigg\}
&=\frac{(12x^2-7)(4x^2+1)-(4x^3-7x)(8x)}{(4x^2+1)^2}\\
&\qquad\qquad\text{by QR with } f(x)=4x^3-7x, f'(x)=12x^2-7,\\
&\qquad\qquad\text{and } g(x)=4x^2+1, g'(x)=8x\\
&=\frac{(48x^4-16x^2-7)-(32x^4-56x^2)}{(4x^2+1)^2}\\
&=\frac{16x^4+40 x^2-7}{(4x^2+1)^2}
\end{align*}
Example2.6.13.Derivative of a cube-root.
In this example, we’ll use a little trickery to find the derivative of \(\root{3}\of{x}\text{.}\) The trickery consists of observing that, by the definition of the cube root,
Example2.6.14.Derivative of a positive rational power of \(x\).
In this example, we’ll use the same trickery as in Example 2.6.13 to find the derivative \(x^{p/q}\) for any two natural numbers \(p\) and \(q\text{.}\) By definition of the \(q^{\rm th}\) root,
That is, \(x^p\) and \(\big(x^{p/q}\big)^q\) are the same function, and so have the same derivative. So we differentiate both of them. We already know that, by Lemma 2.6.9 with \(n=p\text{,}\)
Notice that this has the same form as Lemma 2.6.9, above, except with \(n=\frac{p}{q}\) allowed to be any positive rational number, not just a positive integer.
Example2.6.15.Derivative of \(x^{-m}\).
In this example we’ll use the quotient rule to find the derivative of \(x^{-m}\text{,}\) for any natural number \(m\text{.}\)
By the special case of the quotient rule (Corollary 2.4.6) with \(g(x)=x^m\) and \(g'(x)=mx^{m-1}\)
Again, notice that this has the same form as Lemma 2.6.9, above, except with \(n=-m\) being a negative integer.
Example2.6.16.Derivative of a negative rational power of \(x\).
In this example we’ll use the quotient rule to find the derivative of \(x^{-p/q}\text{,}\) for any pair of natural numbers \(p\) and \(q\text{.}\) By the special case the quotient rule (Corollary 2.4.6) with \(g(x)=x^{\frac{p}{q}}\) and \(g'(x)=\frac{p}{q}x^{\frac{p}{q}-1}\text{,}\)
Note that we have found, in Examples 2.2.2, 2.6.14 and 2.6.16, the derivative of \(x^a\) for any rational number \(a\text{,}\) whether 0, positive, negative, integer or fractional. In all cases, the answer is
Corollary2.6.17.Derivative of \(x^a\).
Let \(a\) be a rational number, then
\begin{equation*}
\diff{}{x} x^a = a x^{a-1}
\end{equation*}
We shall show, in Example 2.10.5, that the formula \(\diff{}{x} x^a = a x^{a-1}\) in fact applies for all real numbers \(a\text{,}\) not just rational numbers.
Back in Example 2.2.9 we computed the derivative of \(\sqrt{x}\) from the definition of the derivative. The above corollary (correctly) gives
As we seen before, the best strategy for dealing with nasty expressions is to break them up into easy pieces. We can think of \(f(x)\) as the five–fold product
Now, to get the derivative \(f(x)\) we use the \(n\)–fold product rule which was developed in Example 2.6.6, together with the special case of the quotient rule (Corollary 2.4.6).
The trick that we used in going from the first line to the second line, namely multiplying term number \(j\) by \(\frac{f_j(x)}{f_j(x)}\) is often useful in simplifying the derivative of a product of many factors 1
Also take a look at “logarithmic differentiation” in Section 2.10.
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Exercises2.6.2Exercises
Exercises — Stage 1
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1.
Spot and correct the error(s) in the following calculation.
