The Other General Differentiation Rules

Section 14.5 The Other General Differentiation Rules

Theorem 14.5.1. The Quotient Rule for Differentiation.

We assume that \(\alpha(x)\text{,}\) \(\beta(x)\text{,}\) and \(f(x)=\frac{\alpha(x)}{\beta(x)}\) are all differentiable functions Assume further that \(\beta(x)\neq0\text{.}\) Then

\begin{equation*} f^\prime(x)=\frac{\beta(x)\alpha^\prime(x)-\alpha(x)\beta^\prime(x)}{\left[\beta(x)\right]^2}. \end{equation*}

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Proving this directly by using limits would be unpleasant, but as we observed in Chapter 4 the Quotient Rule can be viewed as a rearranged version of the Product Rule.

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Problem 14.5.2.

Use the Product Rule to derive the Quotient Rule.

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Hint.

First solve \(f(x)=\frac{\alpha(x)}{\beta(x)}\) for \(\alpha(x)\text{.}\)

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With the Product Rule for Differentiation in place we now have the tools needed to prove the Power Rule for Positive Integer Exponents. The method of proof we outline in the following problem is called Mathematical Induction and it can be used in other contexts as well. In fact, most of the “Find the Pattern” problems in this text require an Induction argument for full rigor.

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Problem 14.5.3. The Power Rule for Positive Integer Exponents.

Assume that \(\alpha(x)=x^n\) for any positive integer \(n\text{.}\)

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(a)

Assume that \(n=1\text{.}\) Use the limit definition to show that \(\alpha^\prime(x) = nx^{n-1}.\) (This says, “The Power Rule holds for \(k=1\text{.}\)”)

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(b)

Now assume that the Power Rule for Positive Integer Exponents holds for \(n=k\text{,}\) where \(k\) is an arbitrary, fixed positive integer. Let \(\beta(x)=x^{k+1}\) and show that \(\beta^\prime(x)=(k+1)x^k.\) (This says, “If the Power Rule holds for \(k\) then it must also hold for \(k+1\text{.}\)”)

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(c)

Do you see how this proves that the Power Rule holds for any positive integer, \(n\text{?}\) Write a short paragraph explaining the logic behind this.

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Part 14.5.3.a and part 14.5.3.b of this problem constitute a rigorous proof of the Power Rule for Positive Integer Exponents by Mathematical Induction, and part 14.5.3.c asks you to explain the underlying logic. Although we didn’t mention it at the time, Mathematical Induction was the underlying idea in Problem 4.3.14 as well.

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With the Power Rule for Positive Integer Exponents in place we can extend it to both negative and rational exponents in the same way we did it in Chapter 4. The following problem is essentially a repeat of Problem 4.3.32 and 4.3.34, using Lagrange’s prime notation, and function notation, rather than differentials.

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Problem 14.5.4. The Power Rule for Rational and Negative Exponents.

(a)

Assume \(n\) is a positive integer and that \(\alpha(x)=x^{-n}\) is differentiable. Show that

\begin{equation*} \alpha^\prime(x) = -nx^{-(n+1)}. \end{equation*}

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Hint.

Rewrite \(\alpha(x)=x^{-n}\) as \(\frac{1}{x^n}\) and use the Quotient Rule for Differentiation and the Power Rule for positive integers.

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(b)

Assume that \(q\) is a non-zero integer and that \(\alpha(x)=x^{1/q}\) is differentiable at \(x\text{.}\) Show that

\begin{equation*} \alpha^\prime(x) = (1/q)x^{(1/q-1)}. \end{equation*}

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Hint.

Rewrite \(\alpha(x)=x^{1/q}\) as

\begin{equation*} \left[\alpha(x)\right]^q=x \end{equation*}

and use the Chain Rule and the Power Rule for positive integers.

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(c)

Assume that \(p\) and \(q\) are integers, \(q\neq0\text{,}\) and that \(\alpha(x)=x^{p/q}\) is differentiable at \(x\text{.}\) Show that

\begin{equation*} \alpha^\prime(x) = (p/q)x^{(p/q-1)}. \end{equation*}

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Hint.

Rewrite \(\alpha(x)=x^{p/q}\) as \(\alpha(x)=\left(x^{1/q}\right)^p\) and use the Chain Rule and part 14.5.4.b.

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Together the previous two problems prove the Power Rule for rational exponents:

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Theorem 14.5.5. The Power Rule for Rational Exponents.

Assume that \(p\) and \(q\) are integers, \(q\neq0\text{,}\) and that \(\alpha(x)=x^{p/q}\) is differentiable at \(x\text{.}\) Then

\begin{equation*} \alpha^\prime(x) = (p/q)x^{(p/q-1)}. \end{equation*}

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In the statement of Theorem 14.5.1 we explicitly assumed that the quotient, \(\frac{\alpha(x)}{\beta(x)}\text{,}\) is differentiable at \(x\text{.}\) This has the effect that the theorem does not necessarily apply to all possible quotients, in the same way that when we add \(\Delta\beta\neq0\) to the statement of the Chain Rule, the theorem applies to fewer compositions. And just like the Chain Rule the functions that Theorem 14.5.1 does not apply to are mostly pathological, and of no use to us right now.

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We added the same assumption to Theorem 14.5.5 for similar reasons.

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DIGRESSION: Are You a Mathematician?

If leaving these theorems incomplete in this way is troubling to you then you are almost certainly a mathematician by temperament. If you haven’t decided on a major yet, consider mathematics. You obviously like it. Why not learn more?

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If you find that you simply don’t care about completing all of the details and you are not majoring in mathematics, congratulations! You’ve made the right choice.

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Problem 14.5.6 will lead you through the steps necessary to prove the Quotient Rule for Differentiation without the assumption that \(\frac{\alpha(x)}{\beta(x)}\) is differentiable. Have fun!

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Problem 14.5.6.

Assume that \(\alpha(x)\) and \(\beta(x)\) are differentiable and that \(\beta(x)\neq0\text{,}\) but we make no assumption about the differentiability of \(f(x)=\frac{\alpha(x)}{\beta(x)}\text{.}\)

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(a)

First prove the special case of the Quotient Rule where \(g(x)=\frac{1}{\beta(x)}\text{.}\)

Use the limit definition to show that \(g^\prime(x)=\limit{h}{0}{\frac{\beta(x)-\beta(x+h)}{h\beta(x)\beta(x+h)}}\text{.}\)
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Now evaluate the limit in part 14.5.6.a to show that \(f^\prime(x)= \frac{-\beta^\prime(x)}{\left[\beta(x)\right]^2}\text{.}\)
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(b)

Use the Product Rule for Differentiation and the Chain Rule (along with the result of part a) to show that \(f(x)=\frac{\alpha(x)}{\beta(x)}\) is differentiable at \(x\) and that

\begin{equation*} \displaystyle f^\prime(x) = \frac{\beta(x)\alpha^\prime(x) - \alpha(x)\beta^\prime(x)}{\left[\beta(x)\right]^2}. \end{equation*}

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Problem 14.5.8 will lead you through the steps necessary to prove the Product Rule for Rational Exponents without the assumption that \(x^{\frac{p}{q}}\) is differentiable. It relies on the result of Problem 14.5.7. Have fun!

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Problem 14.5.7.

To prove Theorem 14.5.5 we will first focus on the special case of

\begin{equation*} \beta(x)=x^{\frac{1}{q}}, \end{equation*}

where \(q\) is a non-negative integer.

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The key to proving this special case is a generalization of the difference of squares formula:

\begin{equation*} (a-b)(a+b)=a^2-b^2. \end{equation*}

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(a)

Show that \((a-b)(a^2+ab+b^2)=a^3-b^3\text{.}\)

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(b)

Show that \((a-b)(a^3+a^2b+ab^2+b^3)=a^4-b^4\text{.}\)

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(c)

Now show that in general

\begin{equation*} (a-b)(a^{q-1}+a^{q-2}b+a^{q-3}b^2+\cdots+ab^{q-2}+b^{q-1})=a^q-b^q\text{.} \end{equation*}

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Problem 14.5.8.

Assume that \(p\) and \(q\) are integers and that \(q\neq0\text{.}\) If we apply Definition 13.2.3 to \(f(x)=x^\frac{1}{q}\text{,}\) we get

\begin{equation*} f^\prime(x)=\limit{h}{0}{\frac{(x+h)^\frac1q-x^\frac1q}{h}}. \end{equation*}

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(a)

Use the substitutions \(a=(x+h)^\frac1q\text{,}\) \(b=x^\frac1q\text{,}\) and part (c) of Problem 14.5.7 to show that

\begin{equation*} f^\prime(x)=\limit{a}{b}{\frac{a-b}{a^q-b^q}}=\frac{1}{qb^{q-1}}. \end{equation*}

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(b)

Substitute \(b=x^{\frac1q}\) into the result of part (a) to obtain

\begin{equation*} f^\prime(x)=\frac1qx^{\frac1q-1}. \end{equation*}

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(c)

Use the Chain Rule to show that for \(\alpha(x)=x^{\frac{p}{q}}\)

\begin{equation*} \alpha^\prime(x)=\frac{p}{q}x^{\frac{p}{q}-1}. \end{equation*}

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END OF DIGRESSION

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