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Section 3.1 Derivatives of Polynomial and Exponential Functions

Our collection of efficiently methods for computing derivatives starts with polynomials and exponential functions. Much as with limits, we do this by first dealing with a few simple functions, and then using rules for handling constant multiples, sums and differences. However, derivatives of products and quotients do not follow the same pattern as with limits, so we leave them until the next section.

Derivatives of Linear Functions.

Since the slope of a straight line \(y=mx+c\) is the constant \(m\text{,}\) it is easy to check that the derivative of \(f(x)=mx+c\) is \(m\text{,}\) for any constants \(m\) and \(c\text{.}\) It is often convenient to write calculations directly with formulas, without naming the functions, so to illustrate several notations:

The two most basic special cases are when the function is a constant \(c\) or just \(x\text{:}\)

\begin{equation*} (c)' = \frac{d}{dx}(c) = 0, \quad (x)' = \frac{d}{dx}(x) = 1. \end{equation*}
Derivatives of Power Functions.

We have already computed the derivatives of a few powers functions like \((x^2)' = 2x\) and \((x^3)' = 3x^2\text{,}\) and these fit a more general pattern:

This also agrees with the results seen above for \(f(x)=x^1 = x\) and \(f(x) = x^0=1\text{.}\)

Note that there is also a “Power Rule for Limits”: from now on, when we simply say “power rule”, we mean this one for derivatives.

To see the pattern that helps us to get the general rule, let us look at \(n=4\text{:}\)

Calculate the derivative of \(f(x)=x^4\text{.}\)

Solution

Use the first formula for the derivative \(f'(a)\text{:}\)

\begin{equation*} f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}. \lim_{x \to a}\frac{x^4 - a^4}{x - a}. \end{equation*}

The numerator vanishes for \(x = a\text{,}\) so it has a factor \(x - a\text{,}\) and in fact the factorization is \(x^4 - a^4 = (x-a)(x^3+x^2 \cdot a + x \cdot a^2 + a^3)\text{.}\) (Check by expanding!)

This gives

\begin{align*} f'(a) \amp = \lim_{x \to a}\frac{(x-a)(x^3+x^2 \cdot a + x \cdot a^2 + a^3)}{x-a}\\ \amp = \lim_{x \to a} (x^3+x^2 \cdot a + x \cdot a^2 + a^3)\\ \amp = 4a^3. \end{align*}

That is, \(f'(x) = (x^4)' = 4x^3\text{,}\) as the Power Rule above says.

Let us try the power rule, using various different notations for derivatives:

For \(f(x) = x^6\text{,}\) find \(f'(x)\text{.}\)

 Answer
\(f'(x) = 6x^5\)

For \(y = x^{1000}\text{,}\) find \(y'\text{.}\)

 Answer
\(y' = 1000x^{999}\)

For \(y = t^{4}\text{,}\) find \(\frac{dy}{dt}\text{.}\)

 Answer
\(\frac{dy}{dt} = 4t^{3}\)

Find \(\frac{d}{dr}(r^3)\text{.}\)

 Answer
\(\frac{d}{dr}(r^3) = 3r^{2}\)
Proof of the Power Rule.

The key step is the factorization

\begin{equation*} x^n-a^n = (x-a)(x^{n-1} + x^{n-2} a + x^{n-3} a^2 + \cdots + xa^{n-2}+ a^{n-1}) \end{equation*}

This can be checked by expanding the right hand side, distributing the left hand factor:

\begin{align*} \amp (x-a)(x^{n-1} + x^{n-2}a + x^{n-3} a^2 + \cdots + xa^{n-2} + a^{n-1})\\ \amp = x(x^{n-1} + x^{n-2} a + x^{n-3} a^2 + \cdots + xa^{n-2} + a^{n-1}) - a(x^{n-1} + x^{n-2} a + x^{n-3} a^2 + \cdots + xa^{n-2} + a^{n-1})\\ \amp = x^n + x^{n-1} a + x^{n-2}a^2 + \cdots + x^{2}a^{n-2} + xa^{n-1} - x^{n-1}a - x^{n-2} a^2 - x^{n-3} a^3 - \cdots - xa^{n-1} - a^n\\ \amp = x^n - a^n \end{align*}

because all the terms in between pair off and cancel out.

Much as with \(x^{4}\text{,}\) the definition of the derivative gives the derivative of \(f(x)=x^n\) at \(x=a\) as

\begin{align*} f'(a) \amp = \lim_{x \to a}\frac{x^n-a^n}{x-a}\\ \amp = \lim_{x \to a}\frac{(x-a)(x^{n-1} + x^{n-2}\cdot a + x^{n-3}\cdot a^2 + \cdots + a^{n-1})}{x-a}\\ \amp = \lim_{x \to a}(x^{n-1} + x^{n-2} \cdot a + x^{n-3} \cdot a^2 + \cdots + a^{n-1})\\ \amp = a^{n-1} + a^{n-2} \cdot a + a^{n-3} \cdot a^2 + \cdots + a^{n-1}\\ \amp = a^{n-1} + a^{n-1} + a^{n-1} + \cdots + a^{n-1} \text{ (} n \text{ copies)}\\ \amp = n a^{n-1}, \end{align*}

so \(f'(x) = (x^n)' = nx^{n-1}\text{,}\) as claimed.

Constant Multiples, Sums and Differences.

As with limits, we can build up polynomials from these power functions using constant multiples, sums and differences. The derivatives of these three basic combinations are as simple as with limits:

Warning: The rules seen so far are the only ones that are as simple and “guessable” as for limits!

Compute the derivative of \(y=x^8 + 12x^5 - 4x^2 + 10x^3 - 6x + 5\text{.}\) Note that this derivative is also a polynomial.

The same approach works for differentiating any polynomial:

All polynomials are differentiable, and their derivatives are polynomials, so the second and higher derivatives also exist.

Find the points on the curve \(y=x^4 - 6x^2 + 4\) where the tangent is horizontal.

Derivatives of Other Power Functions.

Example 3 in Section 2.8 shows that \(\sqrt{x}\) has derivative \({1}/(2 \sqrt{x})\text{.}\) That is, \(\frac{d}{dx}{x^{1/2}} = (1/2)x^{1/2-1}\text{.}\) This fits the power rule, but for power \(1/2\text{,}\) not a positive integer. In fact, this works for all real powers:

This is most easily shown later when we know how to differentiate exponential functions and compositions of functions.

If \(f(x) = {1}/{x^2}\text{,}\) find \(f'(x)\text{.}\)

 Answer
\(f'(x) = {-2}/{x^3}\)

If \(y = \sqrt[3]{x^2}\text{,}\) find \(y'\text{.}\)

 Answer
\(y' = {3}/(2 \sqrt[3]{x})\)

Differentiate \(3/x\text{,}\) and use this to check the slope computed in Example 2 of Section 2.7.

Derivative of the Natural Exponential Function.

In Section 1.5 we defined the number \(e\) so that the slope of \(y=e^x\) at point \((0,1)\) is 1. That is,

\begin{equation*} \lim_{h \to 0}\frac{e^{h}-1}{h} = 1. \end{equation*}

This choice makes the derivative of \(e^x\) simple:

\begin{align*} \frac{d}{dx}e^x \amp = \lim_{h \to 0}\frac{e^{x+h}-e^{x}}{h} = \lim_{h \to 0}\frac{e^x \cdot e^h - e^x}{h} = \lim_{h \to 0}\frac{e^x (e^h - 1)}{h}\\ \amp = e^x \lim_{h \to 0}\frac{e^h - 1}{h}\\ \amp \text{(as } e^x \text{ is a constant as far as this limit is concerned: } h \text{ is the variable!)}\\ \amp = e^x \cdot 1\\ \amp = e^x. \end{align*}

So the natural exponential function is equal to its own derivative:

Thus all exponential functions have a rate of change proportional to their current value.

This fits for example with the simple exponential model of a population whose growth rate is proportional to its current size because the rates of births and deaths are both proportional to current population.

Geometrical Explanation of the Derivative of \(\mathbf{a^x}\).
The result for the derivative of \(a^x\) can be seen graphically by writing \(a=e^{\ln a}\) so that
\begin{equation*} a^x=(e^{\ln a})^x = e^{(\ln a) x}. \end{equation*}
The effect of changing from \(f(x)\) to \(g(x)=f(kx)\) is to compress the graph horizontally by a factor of \(k\text{,}\) increasing the slope at corresponding points by a factor \(k\text{:}\) in terms of derivatives,
\begin{equation*} \frac{d}{dx} f(k x) = k f'(k x) \end{equation*}
Thus, the graph of \(a^x\) is a compression of the graph of \(e^x\) by factor \(\ln a\) and
\begin{equation*} \frac{d}{dx}a^x = (\ln a) e^{(\ln a) x} = (\ln a) a^x.\text{.} \end{equation*}
In Section 3.4 we will see another way to compute this derivative, using a derivative rule for composition of functions.

If \(f(x)=e^x - x\text{,}\) find \(f'(x)\) and \(f''(x)\text{,}\) and then compare the graphs of \(f\) and \(f''\text{.}\)

Answer 1
\(f'(x) = e^x - 1\)
 Answer 2
\(f''(x) = e^x\)

At what point on the curve \(y=e^x\) is the tangent parallel to the line \(y=2x\text{?}\)

Recommended Exercises Recommended Exercises

Study Exercises 6, 9, 16, 23, 31, 34, 40, 46, 48, 49, 55, 77 and 83 from the text.