4.3: The generalized power rule.

In Section 3.5 we proved the Power Rule: for all positive integers, \(n\),

\[[t ^{n} ] ^{\prime} = n ~ t^{n-1}\]

We show here the generalized power rule.

 

Suppose \(n\) is a positive integer and \(u(t)\) is a function that has a derivative for all \(t\). We use the notation

\[(u(t))^{n}=u^{n}(t) . \quad \text { Then } \quad\left[u^{n}(t)\right]^{\prime}=n ~ u^{n-1}(t) \times u^{\prime}(t) \label{4.7}\]

The generalized power rule is used in the following setting. Suppose

\[P(t) = (1 + t ^{2})^3\]

There are two options for computing \(P ^{\prime} (t)\).

Option A. Expand the binomial:

\[\begin{aligned}
P(t) &= (1 + t^{2})^{3}\\
&= 1 + 3t^{2} + 3t^{4} + t^{6}
\end{aligned}\]

Then use the Sum, Constant, Constant Factor, and Power Rules to show that

\[P ^{\prime} (t) = 0 + 6t + 12t ^{3} + 6t ^{5}\]

Option B. Use the Generalized Power Rule with \(u(t) = 1 + t ^{2}\) . Then

\[\begin{aligned}
P(t) &=\left(1+t^{2}\right)^{3} & &=u^{3}(t) \\
P^{\prime}(t) &=3\left(1+t^{2}\right)^{2} \times\left[1+t^{2}\right]^{\prime} & &=3 u^{2}(t) \times u^{\prime}(t) \\
&=3\left(1+t^{2}\right)^{2} \times 2 t & &
\end{aligned}\]

The answers are the same, for

\[3 (1 + t ^{2})^{2} \times 2t = 3(1 + 2t ^{2} + t ^{4}) \times 2t = 6t + 12t ^{3} + 6t ^{5}\]

Option A (expand the binomial) may appear easier than Option B (use the generalized power rule), but the generalized power rule is clearly easier for a problem like

\[\text{Compute} \quad P ^{\prime} (t) \quad \text{for} \quad P(t) = (1 + t^{2})^{10}\]

Expanding \(( 1 + t^{2} ) ^{10}\) into polynomial form is tedious (if you try it you may conclude that it is a worse than tedious). On the other hand, using the generalize power rule

\[P^{\prime}(t)=10\left(1+t^{2}\right)^{9} \times\left[1+t^{2}\right]^{\prime}=10\left(1+t^{2}\right)^{9} \times 2 t\]

A special strength of the generalized power rule is that when \(u\) is positive, Equation \ref{4.7} is valid for all numbers \(n\) (integer, rational, irrational, positive, negative). Thus for \(P(t) = \sqrt{1 + t^{2}}\),

\[\begin{array}{rlr}
P^{\prime}(t) & =\left[\sqrt{1+t^{2}}\right]^{\prime} & \\
& =\left[\left(1+t^{2}\right)^{\frac{1}{2}}\right]^{\prime} & \text{ Change to fractional exponent. } \\
& =\frac{1}{2}\left(1+t^{2}\right)^{\frac{1}{2}-1} \times\left[1+t^{2}\right]^{\prime} & \text { Generalized Power Rule. } \\ & =\frac{1}{2}\left(1+t^{2}\right)^{-\frac{1}{2}} \times 2 t \quad & \text { Sum, Constant, Power Rules }
\end{array}\]

You will prove that when \(u\) is positive Equation \ref{4.7}, \([ u ^{n} (t) ]^{\prime} = n ~ u^{n−1} \times u ^{\prime} (t)\), is valid for \(n\) a negative integer (Exercise 4.3.3) and for \(n\) a rational number (Exercise 4.3.4).

Proof of the Generalized Power Rule. Assume that \(n\) is a positive integer, \(u(t)\) is a function and \(u ^{\prime} (t)\) exists. Then

\[\begin{array}\
&\text{Equations to prove the GPR. See Exercise 4.3.2.}&\\
\left[u^{n}(t)\right]^{\prime}&=\lim _{b \rightarrow t} \frac{u^{n}(b)-u^{n}(t)}{b-t} & (i)\\
&=\lim _{b \rightarrow t} \frac{\left(u^{n-1}(b)+u^{n-2}(b) u(t)+\cdots+u(b) u^{n-2}(t)+u^{n-1}(t)\right) \times(u(b)-u(t))}{b-t} & (ii)\\
&=\lim _{b \rightarrow t}\left(\left(u^{n-1}(b)+u^{n-2}(b) u(t)+\cdots+u(b) u^{n-2}(t)+u^{n-1}(t)\right) \times \frac{(u(b)-u(t)}{b-t}\right) & (iii)\\
&=\lim _{b \rightarrow t}\left(u^{n-1}(b)+u^{n-2}(b) u(t)+\cdots+u(b) u^{n-2}(t)+u^{n-1}(t)\right) \times \lim _{b \rightarrow t} \frac{(u(b)-u(t))}{b-t} & (iv)\\
&=\lim _{b \rightarrow t}\left(u^{n-1}(b)+u^{n-2}(b) u(t)+\cdots+u(b) u^{n-2}(t)+u^{n-1}(t)\right) \times u^{\prime}(t) & (v)\\
&=\left(\lim _{b \rightarrow t} u^{n-1}(b)+\lim _{b \rightarrow t} u^{n-2}(b) u(t)+\cdots+\lim _{b \rightarrow t} u(b) u^{n-2}(t)+\lim _{b \rightarrow t} u^{n-1}(t)\right) \times u^{\prime}(t) & (vi)\\
&=\left(\lim _{b \rightarrow t} u^{n-1}(b)+u(t) \lim _{b \rightarrow t} u^{n-2}(b)+\cdots+u^{n-2}(t) \lim _{b \rightarrow t} u(b)+\lim _{b \rightarrow t} u^{n-1}(t)\right) \times u^{\prime}(t) & (vii)\\
&=\left(\lim _{b \rightarrow t} u^{n-1}(b)+u(t) \lim _{b \rightarrow t} u^{n-2}(b)+\cdots+u^{n-2}(t) \lim _{b \rightarrow t} u(b)+u^{n-1}(t)\right) \times u^{\prime}(t) & (viii)\\
&=\left(\lim _{b \rightarrow t} u^{n-1}(b)+u(t) \lim _{b \rightarrow t} u^{n-2}(b)+\cdots+u^{n-2}(t) \lim _{b \rightarrow t} \times u(t)+u^{n-1}(t)\right) \times u^{\prime}(t) & (ix)\\
&=(\underbrace{u^{n-1}(t)+u^{n-1}(t)+\cdots+u^{n-1}(t)+u^{n-1}(t)}_{n \text { terms }}) \times u^{\prime}(t) & (x)\\
&=n u^{n-1}(t) \times u^{\prime}(t) \quad \text { Whew! }
\end{array} \label{4.8}\]

End of Proof.

Explore 4.3.1 In Explore Figure 4.3.1 is a graph of \(F\) defined by

\[F(x) = x \text{ for } x = 0 \text{ or } x \text{ is the reciprocal of a positive integer.}\]

Only 13 of the infinitely many points of the graph of \(F\) are plotted. What is the graph of \(F ^{\prime}\)? Your vote counts.

Explore Figure 4.3.1 Thirteen of the infinitely many points of the graph of \(y = x\) for \(x = 0\) or \(x\) is the reciprocal of a positive integer.

To demonstrate use of the generalized power rule, we announce a Primary Formula that is proved in Chapter 7.

\[[ \sin {t}] ^{\prime} = \cos {t}, \label{4.9}\]

The derivative of the sine function is the cosine function.

Then, for \((\sin {t}) ^{2} = \sin ^{2} t\),

\[\begin{array} \
[\sin ^{2} t]^{\prime} &= 2 \sin ^{2−1} t \times [ \sin t]^{\prime} &\text{Generalized Power Rule}\\
&= 2 \sin t \times \cos t &\text{Equation } \ref{4.9}\\
\end{array} \label{4.10}\]

Now consider that \(\cos t = \sqrt{1 − \sin ^{2} t}\) for \(0 \leq t \leq \pi /2\).

\[\begin{aligned}\
[\cos t]^{\prime} &=\left[\left(1-\sin ^{2} t\right)^{\frac{1}{2}}\right]^{\prime} & & \text { Definition of } P \\ &=\frac{1}{2}\left(1-\sin ^{2} t\right)^{\frac{1}{2}-1} \times\left[1-\sin ^{2} t\right]^{\prime} & & \text { Generalized Power Rule } \\
&=\frac{1}{2}\left(1-\sin ^{2} t\right)^{\frac{1}{2}-1} \times\left([1]^{\prime}-\left[\sin ^{2} t\right]^{\prime}\right) & & \text { Sum Rule for Derivatives } \\
&=\frac{1}{2} \frac{1}{\sqrt{1-\sin ^{2} t}} \times(0-2 \sin t \cos t) & &[C]^{\prime}=0 \text { and Eq } 4.10 \\
&=-\sin t & & \text { Trigonometric simplification. }
\end{aligned}\]

One might then guess (correctly) that

\[[ \cos t] ^{\prime} = - \sin t \text{ for all } t.\]

Observe the exaggerated ( )’s in the step marked 'Sum Rule for Derivatives.' Students tend to omit writing those parentheses. They may carry them mentally or may loose them. The ( )’s are necessary. Without them, the steps would lead to

\[\begin{aligned}
P^{\prime} &=\frac{1}{2}\left(1-\sin ^{2} t\right)^{\frac{1}{2}-1} \times\left[1-\sin ^{2} t\right]^{\prime} & & \text { Generalized Power Rule } \\
&=\frac{1}{2}\left(1-\sin ^{2} t\right)^{\frac{1}{2}-1} \times[1]^{\prime}-\left[\sin ^{2} t\right]^{\prime} & & \text { Sum Rule for Derivatives } \\
&=\frac{1}{2}\left(1-\sin ^{2} t\right)^{\frac{1}{2}-1} \times 0-2 \sin t \cos t & &[C]^{\prime}=0 \text { and Eq } \ref{4.10} \\
&=-2 \sin t \cos t & & \text { Trigonometric simplification }
\end{aligned}\]

Unfortunately, the answer is incorrect. Always:

First Notice.

Use parentheses, ( )'s, they are cheap.

 

4.3.1 The Power Chain Rule.

The Generalized Power Rule is one of a collection of rules called chain rules and henceforth we will refer to it as the Power Chain Rule. The reason for the word, 'chain' is that the rule is often a 'link' in a 'chain' of steps leading to a derivative. Because of its form

\[[ u(t) ^{n} ] ^{\prime} = n u(t) ^{n-1} \times [ u(t) ]^{\prime} , \]

when the Power Chain Rule is used, there always remains a derivative, \([ u(t) ] ^{\prime}\) , to compute after the Power Chain Rule is used. The power chain rule is never the final step – the final step is always one or more of the Primary Formulas.

For example, compute the derivative of

\[\begin{array}{rlr}\
& y=\frac{1}{1+\sqrt{x+1}}=\left(1+(x+1)^{1 / 2}\right)^{-1} & \\
y^{\prime}= & {\left[\left(1+(x+1)^{1 / 2}\right)^{-1}\right]^{\prime}} & \text { Logical Identity } \\
= & (-1)\left(1+(x+1)^{1 / 2}\right)^{-2} \times\left[1+(x+1)^{1 / 2}\right]^{\prime} & \text { Power Chain Rule } \\
= & (-1)\left(1+(x+1)^{1 / 2}\right)^{-2}\left(0+\left[(x+1)^{1 / 2}\right]^{\prime}\right) & \text { Sum and Constant Rules } \\
= & (-1)\left(1+(x+1)^{1 / 2}\right)^{-2}(1 / 2)(x+1)^{-1 / 2}[x+1]^{\prime} & \text { Power Chain Rule } \\
= & (-1)\left(1+(x+1)^{1 / 2}\right)^{-2}(1 / 2)(x+1)^{-1 / 2}(1+0) & \text { Power and Constant Rules }\\
\end{array}\]

 

Exercises for Section 4.3, The generalized power rule.

Exercise 4.3.1 Compute \(P ^{\prime} (t)\) for

1. \(P(t) = ( 2 + t ^{2} ) ^{4}\)
2. \(P(t) = ( 1 + \sin t) ^3\)
3. \(P(t) = (t ^{4} + 5 )^2\)
4. \(P(t) = ( 6t ^{7} + 5^{4} ) ^9\)
5. \(P(t) = (\frac{t}{8} + t ^{2})^2\)
6. \(P(t) = (t ^{2} + \sin t) ^13\)
7. \(P(t) = (\frac{1}{t} + t)^2\)
8. \(P(t) = (\frac{5}{t} + \frac{t}{3})^2\)
9. \(P(t) = \frac{1}{t + 5}\)
10. \(P(t) = \frac{2}{(1 + t) ^{2}}\)

Exercise 4.3.2 Give reasons to support each of the equality signs labeled (i) − quad − (x) in Equations \ref{4.8} to prove the Generalized Power Rule. Each equality can be justified by reference to algebra, to one of the limit formulas Equations 3.2.26 through 3.2.31 (shown next), or Theorem 4.2.1, The Derivative Requires Continuity, or the definition of the derivative, Equation 3.3.3. Equations 3.2.26 through 3.2.31 are:

\[\begin{array}\
&\text{Eq 3.2.26} \quad \lim _{x \rightarrow a} C=C & \text{Eq 3.2.27} \quad \lim _{x \rightarrow a} x=a\\
&\text{Eq 3.2.28} \quad \lim _{x \rightarrow a} \frac{1}{x}=\frac{1}{a} & \text{Eq 3.2.29} \quad \lim _{x \rightarrow a} C F(x)=C \lim _{x \rightarrow a} F(x)\\
&\text{Eq 3.2.30} \quad \lim _{x \rightarrow a}\left(F_{1}(x)+F_{2}(x)\right)&=\lim _{x \rightarrow a} F_{1}(x)+\lim _{x \rightarrow a} F_{2}(x) \\
&\text{Eq 3.2.31} \quad \lim _{x \rightarrow a}\left(F_{1}(x) \times F_{2}(x)\right)&=\left(\lim _{x \rightarrow a} F_{1}(x)\right) \times\left(\lim _{x \rightarrow a} F_{2}(x)\right)\\
&\text{Eq 3.3.3} \quad F^{\prime}(x)=\lim _{b \rightarrow x} \frac{F(b)-F(x)}{b-x}&\\
\end{array}\]

Exercise 4.3.3 Suppose \(m\) is a positive integer and a function \(u\) has a derivative at \(t\) and that \(u(t) \neq 0\). Give reasons for the equalities \((i) - (vii)\) below that show

\[[u ^{-m(t)}] ^{\prime} = (-m) u ^{-m-1} (t) \times u ^{\prime} (t)\]

\[\[\begin{aligned}
\left[u^{-m}(t)\right]^{\prime}&\stackrel{(i)}{=} \lim _{b \rightarrow 0} \frac{u^{-m}(b)-u^{-m}(t)}{b-t}\\
&=\lim _{b \rightarrow t} \frac{\frac{1}{u^{m}(b)}-\frac{1}{u^{m}(t)}}{b-t}\\
&=\lim _{b \rightarrow t} \frac{u^{m}(t)-u^{m}(b)}{u^{m}(b) \times u^{m}(t)} \times \frac{1}{b-t}\\
&\stackrel{(i i)}{=} \quad-\lim _{b \rightarrow t} \frac{u^{m-1}(t)+u^{m-2}(t) u(b)+\cdots+u(t) u^{m-2}(b)+u^{m-1}(b)}{u^{m}(b) \times u^{m}(t)} \times \frac{u(b)-u(t)}{b-t}\\
&\stackrel{(i i i)}{=} \quad-\frac{\lim _{b \rightarrow t}\left(u^{m-1}(t)+u^{m-2}(t) u(b)+\cdots u(t) u^{m-2}(b)+u^{m-1}(b)\right)}{\lim _{b \rightarrow t} u^{m}(b) \times u^{m}(t)} \times \lim _{b \rightarrow t} \frac{u(b)-u(t)}{b-t}\\
&\stackrel{(i v)}{=} \quad-\frac{\lim _{b \rightarrow t}\left(u^{m-1}(t)+u^{m-2}(t) u(b)+\cdots u(t) u^{m-2}(b)+u^{m-1}(b)\right)}{u^{m}(t) \times u^{m}(t)} \times \lim _{b \rightarrow t} \frac{u(b)-u(t)}{b-t}\\
&\stackrel{(v)}{=} \quad-\frac{\lim _{b \rightarrow t}\left(u^{m-1}(t)+u^{m-2}(t) u(b)+\cdots u(t) u^{m-2}(b)+u^{m-1}(b)\right)}{u^{m}(t) \times u^{m}(t)} \times u^{\prime}(t)\\
&\stackrel{(v i)}{=} \quad-\frac{u^{m-1}(t)+u^{m-1}(t)+\cdots u^{m-1}(t)+u^{m-1}(t)}{u^{m}(t) \times u^{m}(t)} \times u^{\prime}(t)\\
&\stackrel{(v i i)}{=} \quad(-m) u^{-m-1}(t) \times u^{\prime}(t)
\end{aligned}\]\]

Exercise 4.3.4 Suppose \(p\) and \(q\) are integers and \(u\) is a positive function that has a derivative at all numbers \(t\). Assume that \([u ^{\frac{p}{q}} (t)]^{\prime}\) exists. Give reasons for the steps \((i) − (iv)\) below that show

\[[u ^{\frac{p}{q}} (t)]^{\prime} = \frac{p}{q} u ^{\frac{p}{q} -1} (t) \times u ^{\prime} (t).\]

Let

\[v(t) = u ^{\frac{p}{q}} (t).\]

Then

\[\begin{aligned}
v^{q}(t)&=u^{p}(t) &(i)\\
\left[v^{q}(t)\right]^{\prime}&=\left[u^{p}(t)\right]^{\prime}\\
q v^{q-1}(t) \times v^{\prime}(t)&=p u^{p-1}(t) \times u^{\prime}(t) &(ii)\\
v^{\prime}(t) &=\frac{p}{q} \frac{u^{p-1}(t)}{\left(u^{\frac{p}{q}}\right)^{q-1}} \times u^{\prime}(t) &(iii)\\
\left[u^{\frac{p}{q}}(t)\right]^{\prime} &=\frac{p}{q} u^{\frac{p}{q}-1}(t) \times u^{\prime}(t) &(iv)\\
\end{aligned}\]