If we give a function the name f and tell it to operate on a variable x, then f is an nth-degree polynomial function if and only if it can be written in this form:
f (x ) =anx n+an−1x n−1+an−2x n−2+ · · · +a1x+a0
where each addend is called a term, the subscripted letters a1,a2,a3, . . . , and an represent real numbers called the coefficients of the terms, the subscripted letter a0 represents the stand-alone constant or constant term, and n is a positive integer.
Summing the Powers 79
How it looks
Here are some examples of polynomial functions. They are of degree 4, 5, 7, and 11, respectively:
f1(x ) = 6x4− 3x 3+ 3x 2+ 2x+ 5 f2(x ) = 3x5− 4x 3 f3(x ) = −x7− 5x4+ 3x 3−x 2− 29
f4(x ) = −4x11
In all but the first of these functions, some of the coefficients are equal to 0. The coefficient an
by which x n is multiplied, called the leading coefficient, can’t be 0 in an nth-degree polynomial function. If we set an= 0 in a polynomial function f (x ), we end up with
f (x ) = 0x n+an−1x n−1+an−2x n−2+ · · · +a1x+a0
This expression is not technically wrong, but it contains a useless term. It’s really a single-variable polynomial function of degree n− 1:
f (x ) =an−1x n−1+an−2x n−2+ · · · +a1x+a0
We assume, of course, that an−1≠ 0!
Break it down
Look closely at the general form for an nth-degree polynomial function. Then look at each of the four examples above. Every one of these functions has something in common. They’re all sums of monomial power functions. For example,
g (x ) = 6x4− 3x 3+ 3x 2+ 2x+ 5
is the sum of a monomial fourth-degree (or quartic) function, a monomial cubic function, a monomial quadratic function, a monomial linear function, and a constant function. Let’s give each of these functions the name of g with a subscript indicating its degree, like this:
g4(x ) = 6x4
g3(x ) = −3x 3
g2(x ) = 3x 2 g1(x ) = 2x g0 (x ) = 5 We can write g (x ) as the sum of these:
g (x ) =g4(x ) +g3(x ) +g2(x ) +g1(x ) +g0(x )
We’ve learned that the derivative of a sum is always equal to the sum of the derivatives. We’ve also learned how to differentiate any function that takes the variable to a nonnegative integer
power and then multiplies it by a constant. Therefore, we have the tools to differentiate any polynomial function.
Differentiate each term
Once we’ve broken a polynomial function down into its terms, we can use the power rule to differentiate each one of them. In the above situation, we get these derivatives:
g4′ (x ) = 24x 3 g3′ (x ) = −9x 2 g2′ (x ) = 6x g1′ (x ) = 2 g0′ (x ) = 0
Put it back together
To do the final step in working out the derivative of the polynomial function, we add the derivatives of the individual terms together in order of highest-to-lowest powers:
g′(x ) = g4′ (x ) +g3′ (x ) +g2′ (x ) +g1′ (x ) +g0′ (x ) = 24x 3− 9x 2+ 6x+ 2 + 0 = 24x 3− 9x 2+ 6x+ 2
Are you confused?
Do you wonder what happens if we try to find the derivative of a function that raises the variable to a negative-integer power such as −5, or to a non-integer rational power such as 3/5 or −3.7, or even to an irrational power such as π or the square root of 2? What about functions that don’t involve exponents, such as the sine or cosine? We’ll explore the derivatives of some such functions in Chap. 7.
Here’s a challenge!
The sum rule for two derivatives, stated earlier in this chapter, says that if f1 and f2 are differentiable func-tions of the same variable, then
( f1+f2)′ =f1′ +f2′
We’ve seen some examples of this rule “in action,” but we haven’t proved it yet. Now is the time!
Solution
Before we begin the proof, let’s make four changes to the notation:
• Include the name of the variable (we’ll use x )
• Change the names of the functions to f and g
• Write δ (lowercase Greek delta) instead of Δx to represent a shrinking increment
• Leave out “δ → 0” beneath “Lim” (but remember that it’s implied)
Summing the Powers 81
These changes will make the expressions clearer than they would be otherwise. They’ll also help you get used to some of the alternative ways that things can be written down in calculus. We want to prove that if f and g are differentiable functions of x, then
( f+g)′ (x ) =f ′ (x ) +g′ (x )
According to the definition at the end of Chap. 3, the derivative of our sum function, written the “new way,” is
( f+g)′ (x ) = Lim [( f+g) (x+ δ)− ( f+g) (x )] / δ We can rewrite this as
Lim {[ f (x+ δ)+g (x+ δ)]− [ f (x ) +g (x )]} / δ
which can be rearranged to
Lim [ f (x+ δ)+g (x+ δ)−f (x ) −g (x )] / δ and further to
Lim {[ f (x+ δ)−f (x )] + [ g (x+ δ)−g (x )]} / δ
and still further to
Lim [ f (x+ δ)−f (x )] / δ + [ g (x+ δ)−g (x )] / δ
In Chap. 2, we learned that the limit of a sum is equal to the sum of the limits. Using that rule, we can split the above limit into a sum of two limits, getting
Lim [ f (x+ δ)−f (x )] / δ + Lim [ g (x+ δ)−g (x )] / δ
To finish, we can apply the definition of the derivative “backward” to both of these limits and rewrite the above expression as
f ′(x ) +g′(x )
We started out with ( f+g)′ (x ) and ended up with f ′(x ) +g′(x ), showing that these two expressions are equivalent. Mission accomplished!
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. A. The solutions in the appendix may not represent the only way a problem can be figured out. If you think you can solve a particular problem in a quicker or better way than you see there, by all means try it!
1. Determine the derivatives of the following functions:
(a) f (x ) = −8x5 (b) g (z)= 12z7 (c) h (t)= −21t21
2. Consider the function f3(x ) =x 3 in the interval 0 ≤x≤ 1. Find x 0 in this interval, accurate to three decimal places, such that f3′ (x 0)= 1. (It’s okay to use a calculator.) 3. Consider the function f5(x ) =x5 in the interval 0 ≤x≤ 1. Find x 0 in this interval,
accurate to three decimal places, such that f5′ (x 0)= 1. (It’s okay to use a calculator.) 4. Consider the infinite set of functions fn(x ) =x n in the interval 0 ≤x≤ 1, where n is an
integer larger than or equal to 2. Find a general expression for x 0 in this interval, such thatfn′ (x 0)= 1. (A calculator is useless here!)
5. Sketch a multi-curve graph that shows fn(x ) =x n in the interval 0 ≤x≤ 1 for n= 3, n= 4, and n= 5. On each curve, show the point x 0 at which fn′ (x 0)= 1 as a solid dot.
To plot the points, use the values obtained in the solutions of problems we’ve already solved in this chapter. Draw lines tangent to the curves at these points.
6. Consider again the set of functions fn(x ) =x n, where n is an integer larger than or equal to 2. Find a general expression for the value of fn′ (1).
7. Consider (yet again!) the set of functions fn(x ) =x n, where n is an integer larger than or equal to 2. Find a general expression for the value of fn′ (0).
8. Find the derivative of the function
f (x ) = 8x7+ 4x6− 3x5+x4+x 2− 3
9. Find the derivative of the function
f (x ) =a5x5+a4x4+a3x 3+a2x 2+a1x+a0
where a5,a4,a3,a2,a1, and a0 are real numbers, and a5≠ 0.
10. Write an expression for the derivative of the general polynomial function f (x ) =anx n+an−1x n−1+an−2x n−2+ · · · +a2x 2+a1x+a0
where an,an−1,an−2, . . . , a1, and a0 are real numbers, and an≠ 0.
Practice Exercises 83
84