4.1: Rules of Differentiation

Learning Objectives

1. Express the power rule (Table 4.1) and be prepared to apply it to both derivatives and antiderivatives of power functions and polynomials.
2. Explain what is meant by the statement that "the derivative is a linear operation".
3. Describe the concept of an antiderivative and why it is defined only up to some constant.
4. Express the product and quotient rules and be able to apply these to calculating derivatives of products and of rational functions. The derivative of power functions: the power rule

The derivative of power functions: the power rule

We have already computed the derivatives of several of the power functions. See Example 3.5 for \(y=x^{0}=1\) and Example 3.6 for \(y=x^{1}\). See also Example \(2.10\) for \(y=x^{2}\) and Example \(3.12\) for \(y=x^{3}\). We gather these results in Table 4.1. From the table, we observe that the derivative of a power function is also a power function: the original power becomes a coefficient and the new power is reduced by 1 . We refer to this pattern as the power rule of differentiation.

Table 4.1: The Power Rule of differentiation states that the derivative of the power function \(y=x^{n}\) is \(n x^{n-1}\). For now, we have established this result for integer \(n\). Later, we find this result holds for other powers that are not integer.

Function	Derivative
\(f(x)\)	\(f^{\prime}(x)\)
1	0
\(x\)	1
\(x^{2}\)	\(2 x\)
\(x^{3}\)	\(3 x^{2}\)
\(\vdots\)	\(\vdots\)
\(x^{n}\)	\(n x^{n-1}\)
\(x^{n / m}\)	\((n / m) x^{(n / m)-1}\)

We can show that this rule applies for any power function of the form \(y=f(x)=x^{n}\) where \(n\) is an integer power. The calculation is essentially the same as examples illustrated in a previous chapter, but the step of expanding the binomial \((x+h)^{n}\) entails lengthier algebra. (Such expansion contains terms of the form \(x^{n-k} h^{k}\) multiplied by binomial coefficients, and we include the details in Appendix E.) From now on, we simply use this result for any power function with integer powers (rather than calculating the derivative using its definition).

Example 4.1

Find the equation of the tangent line to the graph of the power function \(y=f(x)=4 x^{5}\) at \(x=1\), and determine the \(y\)-intercept of that tangent line.

Solution

Using the power rule, the derivative of the function is

\[f^{\prime}(x)=20 x^{4} . \nonumber \]

At the point \(x=1\), we have \(d y / d x=f^{\prime}(1)=20\) and \(y=f(1)=4\). This means that the tangent line goes through the point \((1,4)\) and has slope 20. Thus, its equation is

\[\frac{y-4}{x-1}=20 \quad \Rightarrow \quad y=4+20(x-1)=20 x-16 \nonumber \]

Letting \(x=0\) in the equation of the tangent line, we find that the \(y\)-intercept of line is \(y=-16\).

Mastered Material Check

1. Verify that the point \((1,4)\) satisfies the equation \(y=20 x-16\). Next, we find that the result for derivatives of power functions can be extended to derivatives of polynomials, using simple properties of the derivative.

Example 4.2 (Energy loss and Earth’s temperature)

In Section 1.3, we studied the energy balance on Earth. According to Equation (1.5), the rate of loss of energy from the surface of the Earth depends on its temperature according to the rule

\[E_{\text {out }}(T)=4 \pi r^{2} \varepsilon \sigma T^{4} . \nonumber \]

Calculate the rate of change of this outgoing energy with respect to the temperature \(T\).

Solution

The quantities \(\pi, \varepsilon, r\) are constants for this problem. Hence the rate of change (’derivative’) of energy with respect to \(T\), denoted \(E_{\text {out }}^{\prime}(T)\) is

\[E_{\text {out }}^{\prime}(T)=\left(4 \pi r^{2} \varepsilon \sigma\right) \cdot 4 T^{3}=\left(16 \pi r^{2} \varepsilon \sigma\right) T^{3} . \nonumber \]

Next, we find that the result for derivatives of power functions can be extended to derivatives of polynomials, using simple properties of the derivative.

The derivative is a linear operation

The derivative satisfies several convenient properties, among them:

1. the derivative of a sum of two functions is the same as the sum of the derivatives; and
2. a constant multiplying a function can be brought outside the derivative.

We summarize these rules:

The derivative is a linear operation, that is:

\[\frac{d}{d x}(f(x)+g(x))=\frac{d f}{d x}+\frac{d g}{d x}\]

\[\frac{d}{d x} C f(x)=C \frac{d f}{d x}\]

In general, a linear operation \(L\) is a rule or process that satisfies two properties:

1. \(L[f+g]=L[f]+L[g]\) and
2. \(L[c f]=c L[f]\),

where \(f, g\) are objects (such as functions, vectors, etc.) on which \(L\) acts, and \(c\) is a constant multiplier. We refer to Equations (4.1) and (4.2) as the linearity properties of the derivative.

Mastered Material Check

2. Verify Equations (4.1.1) and (4.1.2) hold for \(f(x)=x^{3}, g(x)=x^{2}\) and \(C=4\).
3. Give an example which shows squaring is not a linear operation. (Observe that each term consists of the coefficient times the derivative of a power function. The constant term \(a_{0}\) has disappeared since the derivative of any constant is zero.) The derivative, \(p^{\prime}(x)\), is a function in its own right, and a polynomial as well. Its degree, \(n-1\), is one less than that of \(p(x)\). In view of this observation, we could ask: what is the derivative of the derivative?

The derivative of a polynomial

Using the linearity of the derivative, we can extend our differentiation power rule to compute the derivative of any polynomial. Recall that polynomials are sums of power functions multiplied by constants. A polynomial of degree \(n\) has the form

\[p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots a_{1} x+a_{0}, \nonumber \]

where the coefficients, \(a_{i}\) are constant and \(n\) is an integer. Hence, the derivative of a polynomial is just the sum of derivatives of power functions (multiplied by constants). Formally, the derivative of Equation (4.3) is

\[p^{\prime}(x)=\frac{d y}{d x}=a_{n} \cdot n x^{n-1}+a_{n-1} \cdot(n-1) x^{n-2}+\cdots+a_{1} . \nonumber \]

(Observe that each term consists of the coefficient times the derivative of a power function. The constant term \(a_0\) has disappeared since the derivative of any constant is zero.) The derivative, \(p^{\prime}(x)\), is a function in its own right, and a polynomial as well. Its degree, \(n−1\), is one less than that of \(p(x)\). In view of this observation, we could ask: what is the derivative of the derivative?

Notation: we henceforth refer to the "derivative of the derivative" as a second derivative, written in the notation \(p^{\prime \prime}(x)\) or, equivalently \(\frac{d^{2} p}{d x^{2}}\).

Using the same rules, we can compute, obtaining

\[p^{\prime \prime}(x)=\frac{d^{2} p}{d x^{2}}=a_{n} n(n-1) x^{n-2}+a_{n-1}(n-1)(n-2) x^{n-3}+\cdots+a_{2} . \nonumber \]

Mastered Material Check

4. Are their other notations for the second derivative of \(p(x)\) that you might expect?

The following examples should be used for practice.

Example 4.3

Find the first and second derivatives of the function

(a) \(y=f(x)=2 x^{5}+3 x^{4}+x^{3}-5 x^{2}+x-2\) with respect to \(x\) and

(b) \(y=f(t)=A t^{3}+B t^{2}+C t+D\) with respect to \(t\).

Solution

We obtain the results

(a) \(f^{\prime}(x)=10 x^{4}+12 x^{3}+3 x^{2}-10 x+1\) and \(f^{\prime \prime}(x)=40 x^{3}+36 x^{2}+6 x-10\).

(b) \(f^{\prime}(t)=3 A t^{2}+2 B t+C\) and \(f^{\prime \prime}(t)=6 A t+2 B\).

In (b) the independent variable is \(t\), but, of course, the rules of differentiation are the same.

Antiderivatives of power functions and polynomials

Given a derivative, we can ask "what function was differentiated to lead to this result?" This reverse process is termed antidifferentiation, and the function we seek is then called an antiderivative. Antidifferentiation reverses the operation of differentiation. We ask, for example, which function has as its derivative

\[y^{\prime}(t)=A t^{n} ? \]

The original function, \(y(t)\), should have a power higher by 1 (of the form \(t^{n+1}\) ), but the "guess" \(y_{\text {guess }}=A t^{n+1}\) is not quite right, since differentiation results in \(A(n+1) t^{n}\). To fix this, we revise the "guess" to

\[y(t)=A \frac{1}{(n+1)} t^{n+1} . \]

Mastered Material Check

5. Check that Equation (4.1.4) is an antiderivative of Equation (4.1.3) by differentiating Equation (4.1.4).

Question. Is this the only function that has the desired property? No, there are other functions whose derivatives are the same. For example, consider adding an arbitrary constant \(C\) to the function in Equation (4.1.4) and note that we obtain the same derivative (since the derivative of the constant is zero). We summarize our findings:

\[\text { The antiderivative of } y^{\prime}(t)=A t^{n} \quad \text { is } \quad y(t)=A \frac{1}{(n+1)} t^{n+1}+C \text {. } \nonumber \]

We also note a similar result that holds for functions in general:

Given a function, \(f(x)\) we can only determine its antiderivative up to some (additive) constant.

We can extend the same ideas to finding the antiderivative of a polynomial.

Example 4.4 (Antiderivative of a polynomial)

Find an antiderivative of the polynomial \(y^{\prime}(t)=A t^{2}+B t+C\).

Solution

Since differentiation is a linear operation, we can construct the antiderivative by antidifferentiating each of the component power functions. Applying Equation (4.8) to the components we get,

\[y(t)=A \frac{1}{3} t^{3}+B \frac{1}{2} t^{2}+C t+D, \nonumber \]

where \(D\) is an arbitrary constant. We see that the antiderivative of a polynomial is another polynomial whose degree is higher by 1.

Example 4.5

The second derivative of some function is \(y^{\prime \prime}(t)=c_{1} t+c_{2}\). Find a function \(y(t)\) for which this is true.

Solution

The above polynomial has degree 1. Evidently, this function resulted by taking the derivative of \(y^{\prime}(t)\), which had to be a polynomial of degree 2 . We can check that either \(y^{\prime}(t)=\frac{c_{1}}{2} t^{2}+c_{2} t\), or \(y^{\prime}(t)=\frac{c_{1}}{2} t^{2}+c_{2} t+\) \(c_{3}\) (for any constant \(c_{3}\) ) could work. In turn, the function \(y(t)\) had to be a polynomial of degree 3 . One such function is

\[y(t)=\frac{c_{1}}{6} t^{3}+\frac{c_{2}}{2} t^{2}+c_{3} t+c_{4}, \nonumber \]

where \(c_{4}\) is any constant. In short, the relationship is:

\[\text { for differentiation } \quad y(t) \rightarrow y^{\prime}(t) \rightarrow y^{\prime \prime}(t) \text {, } \nonumber \]

whereas

\[\text { for antidifferentiation } \quad y^{\prime \prime}(t) \rightarrow y^{\prime}(t) \rightarrow y(t) \text {. } \nonumber \]

These results are used in applications to acceleration, velocity, and displacement of a moving object in Section 4.2.

Mastered Material Check

6. What is an example of a constant that is not additive?
7. Verify the solution to Example \(4.4\) by differentiating.
8. In Example 4.5, identify all:
   1. constants
   2. dependent variables
   3. independent variables

Product and quotient rules for derivatives

So far, using the power rule, and linearity of the derivative (Section 4.1), we calculated derivatives of polynomials. Here we state without proof (see Appendix E), two other rules of differentiation.

The Product Rule

If \(f(x)\) and \(g(x)\) are two functions, each differentiable in the domain of interest, then

\[\frac{d[f(x) g(x)]}{d x}=\frac{d f(x)}{d x} g(x)+\frac{d g(x)}{d x} f(x) . \nonumber \]

Another notation for this rule is

\[[f(x) g(x)]^{\prime}=f^{\prime}(x) g(x)+g^{\prime}(x) f(x) . \nonumber \]

Example 4.6

Find the derivative of the product of the two functions \(f(x)=x\) and \(g(x)=1+x\).

Solution

Using the product rule leads to

\[\begin{array}{rlr} \frac{d[f(x) g(x)]}{d x}=\frac{d[x(1+x)]}{d x} & =\quad \frac{d[x]}{d x} \cdot(1+x)+\frac{d[(1+x)]}{d x} \cdot x \\ & =\quad 1 \cdot(1+x)+1 \cdot x=2 x+1 . \end{array} \nonumber \]

Mastered Material Check

9. What does ’domain of interest’ mean?
10. What is the domain of the function \(f(x)=\frac{1}{x}\) ?
11. Verify Example \(4.6\) by noting \(f(x) g(x)=x(1+x)=x+x^{2}\) and differentiating.

The Quotient Rule

If \(f(x)\) and \(g(x)\) are two functions, each differentiable in the domain of interest, then

\[\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{\frac{d f(x)}{d x} g(x)-\frac{d g(x)}{d x} f(x)}{[g(x)]^{2}} . \nonumber \]

We can also write this in the form

\[\left[\frac{f(x)}{g(x)}\right]^{\prime}=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{[g(x)]^{2}} . \nonumber \]

Example 4.7

Find the derivative of the function \(y=a x^{-n}=a / x^{n}\) where \(a\) is a constant and \(n\) is a positive integer.

Solution

We can rewrite this as the quotient of the two functions \(f(x)=a\) and \(g(x)=x^{n}\). Then \(y=f(x) / g(x)\) so, using the quotient rule leads to the derivative

\[\begin{aligned} \frac{d y}{d x}=\frac{f^{\prime}(x) g(x)-g^{\prime}(x) f(x)}{[g(x)]^{2}} & =\frac{0 \cdot x^{n}-\left(n x^{n-1}\right) \cdot a}{\left(x^{n}\right)^{2}} \\ & =-\frac{a n x^{n-1}}{x^{2 n}}=a(-n) x^{-n-1} \end{aligned} \nonumber \]

This calculation shows that the power rule of differentiation holds for negative integer powers.

Featured Problem 4.1 (Dynamics of actin in the cell): Actin is a structural protein that forms long filaments and networks in living cells. The actin network is continually assembling from small components (actin monomers) and disassembling back again. To study this process, scientists attach fluorescent markers to actin, and watch the fluorescence intensity change over time. In one experiment, both red and green fluorescent labels were used. The green label fluoresces only after it is activated by a pulse of light, whereas the red fluorescent protein is continually active.

It was noted that the red and green fluorescence intensities \((R, G)\) satisfied the following relationships:

\[\frac{d R}{d t}=(a-b) R, \quad \frac{d G}{d t}=-b G \nonumber \]

where \(a, b\) are constants that characterize the rate of assembly and disassembly ("breakup") of actin. Show that the derivative of the ratio, \(\frac{d(R / G)}{d t}\), can be expressed in terms of the ratio \(R / G\).

Featured Problem \(4.2\) (Derivative of the Beverton-Holt function): The Beverton-Holt model for fish population growth was discussed in Featured Problem 1.1. A function relating the population of fish this year, \(y\) to the population of fish in the previous year \(x\) was

\[y=f(x)=k_{1} \frac{x}{\left(1+k_{2} x\right)} \nonumber \]

(where we have simplified the notation, \(x=N_{0}, y=N_{1}\) of Equation 1.9. How sensitive is this year’s population to slight changes in last year’s population? Compute the derivative \(d y / d x\) to answer this question.

The chain rule states that the derivative of this new function with respect to \(x\) is the product of derivatives of the individual functions. Such relationships between a function of time and its own derivative are examples of differential equations, a topic we revisit in Chapter 11.

A preview of the chain rule

We give a brief preview of the chain rule, an important rule discussed in detail in Chapter 8 . This rule extends the rules of differentiation to composite functions, that is, functions made up of applying a sequence of operations one after the other. For example, the two functions

\[f(u)=u^{10}, \quad u=g(x)=3 x^{2}+1\]

could be applied one after the other to lead to the new composite function

\[f(g(x))=\left(3 x^{2}+1\right)^{10} . \nonumber \]

The Chain Rule

If \(y=f(z)\) and \(z=g(x)\) are two functions, then the derivative of the composite function \(f(g(x))\) is

\[\frac{d y}{d x}=\frac{d f}{d z} \frac{d z}{d x} . \nonumber \]

Example 4.8

Use the chain rule to differentiate the composite function \(y=f(g(x))=\left(3 x^{2}+1\right)^{10}\).

Solution

The functions being composed are the same as in Equation (4.1.5).

Applying the chain rule gives:

\[\frac{d y}{d x}=\frac{d f}{d u} \frac{d u}{d x}=10 u^{9} \cdot 6 x=60\left(3 x^{2}+1\right)^{9} x . \nonumber \]

The details of how to use the chain rule, and many applications are postponed to Chapter 8.

The power rule for fractional powers

Using the definition of the derivative, we have already shown that the derivative of \(\sqrt{x}\) is \(y^{\prime}(x)=\frac{1}{2 \sqrt{x}}\) (see Exercise 2.8). We restate this result using fractional power notation. Recall that \(\sqrt{x}=x^{1 / 2}\).

The derivative of \(y=\sqrt{x}\) is \(y^{\prime}(x)=\frac{1}{2} x^{-1 / 2}\).

This idea can be generalized to any fractional power. Indeed, we state here a result (to be demonstrated in Chapter 9).

Derivative of Fractional-Power Function

The derivative of \(y=f(x)=x^{m / n} \quad\) is \(\quad \frac{d y}{d x}=\frac{m}{n} x^{\left(\frac{m}{n}-1\right)}\).

Notice that this appears in the last row of Table 4.1.

Example 4.9 (Energy loss and Earth’s temperature, revisited)

In Example 4.2, we calculated the rate of change of energy lost per unit change in the Earth’s temperature based on Equation (1.5). Find the rate of change of Earth’s temperature per unit energy loss based on the same equation.

Solution

We are asked to find \(d T / d E_{\text {out }}\). We first rewrite the relationship to express \(T\) as a function of \(E_{\text {out }}\). To do so, we solve for \(T\) in Equation (1.5), obtaining

\[T=\left(\frac{E_{\text {out }}}{4 \pi r^{2} \varepsilon \sigma}\right)^{1 / 4}=\left(\frac{1}{4 \pi r^{2} \varepsilon \sigma}\right)^{1 / 4} E_{\text {out }}^{1 / 4}=K E_{\text {out }}^{1 / 4} . \nonumber \]

The first term is a constant, and we use the rule for a derivative of a fractional power to compute that

\[\frac{d T}{d E_{\text {out }}}=\left(\frac{1}{4 \pi r^{2} \varepsilon \sigma}\right)^{1 / 4} \cdot \frac{1}{4} \cdot E_{\text {out }}^{-3 / 4} . \nonumber \]

Note: in Chapter 9 we will show that implicit differentiation can be used to find the desired derivative without first solving for the variable of interest.