The Power Rule, the Basic Rules of Differentiation, and the Derivatives of Polynomials

Section 4.1 Problems

4.2 The Power Rule, the Basic Rules of Differentiation, and the Derivatives of Polynomials

In this section, we will begin a systematic treatment of the computation of derivatives.

Knowing how to differentiate is fundamental to your understanding of the rest of the course. Although computer software is now available to compute derivatives of many functions (such asy=cxⁿory=e^sinx), it is nonetheless important that you master the techniques of differentiation.

The power rule is the simplest of the differentiation rules. It allows us to compute the derivative of a function of the formy=xⁿ, wherenis a positive integer.

Power Rule Letnbe a positive integer; then d

d x(xⁿ)=nxⁿ⁻¹

We found the rule for the constant function f(x)=ain the previous section.

If f(x)is the constant function f(x)=a, then d

d x f(x)=⁰

We prove the power rule first forn=2—that is, for f(x)=x²(Figure 4.20). In Subsection 4.1.1, we computed the derivative ofy= x²atx =1. In this section, we compute the difference quotient^f_x at any arbitraryx:

f

x = f(x+h)− f(x)

h = (x+h)²−x² h

x h x x

y

h

f(x h) f(x) (x h)² x² (x h, (x h)²)

(x, x²)

Secant line x²

Figure 4.20 The slope of the secant line through(x,x²) and(x+h, (x+h)²)is^(x+h)_h^2−x2.

Using the expansion(x+h)² =x²+²x h+h², we find that f

x = x²+²x h+h²−x²

h = ²x h+h²

h =²x+h (4.5)

after cancelinghin both the numerator and the denominator. To find the derivative, we need to leth→0:

f(x)= ^lim

x→0

f x = ^lim

h→0(²x+h)=²x

This sequence of steps proves the power rule forn = 2. The proof of the rule for other positive integers ofn is conceptually no different from the casen = 2, but it gets algebraically much more involved. For generaln, we need the expansion of (x+h)ⁿ, given by thebinomial theorem, which we will not prove.

Binomial Theorem Ifnis a positive integer, then (x+y)ⁿ =xⁿ+nxⁿ⁻¹y+ n(n−¹)

2·¹ xⁿ⁻²y² + n(n−¹)(n−²)

3·²·¹ xⁿ⁻³y³ + · · · +n(n−1)· · ·(n−k+1)

k(k−¹)· · ·²·¹ x^n−ky^k + · · · +nx yⁿ⁻¹+yⁿ

The expansion of(x+y)ⁿis thus a sum of terms of the form C_n,kx^n−ky^k, k=0,1, . . . ,n

whereCn,kis a coefficient that depends onnandk. The exact form of the coefficients Cn,k will not be important in the proof of the power rule, except for the two terms Cn,0 =^{1 and}Cn,1=n, which are the coefficients forxⁿ andxⁿ⁻¹y, respectively.

Proof of the Power Rule

We use the binomial theorem to expand and then compute the difference in the numerator of the difference quotient:

f = f(x+h)− f(x)=(x+h)ⁿ−xⁿ

=(C_n,0xⁿ+C_n,1xⁿ⁻¹h+C_n,2xⁿ⁻²h²+C_n,3xⁿ⁻³h³ + · · · +Cn,n−1x hⁿ⁻¹+Cn,nhⁿ)−xⁿ

SinceC_n,0=^{1, the}xⁿ terms cancel. We can then factorhout of the remaining terms and find that

f(x+h)− f(x)=h

C_n,1xⁿ⁻¹+C_n,2xⁿ⁻²h+C_n,3xⁿ⁻³h²+ · · · +C_n,nhⁿ⁻¹ When we divide byhand leth→0, we obtain

f(x)= ^lim

h→0

f(x+h)− f(x) h

= ^lim

h→0

C_n,1xⁿ⁻¹+C_n,2xⁿ⁻²h+C_n,3xⁿ⁻³h²+ · · · +C_n,nhⁿ⁻¹

All terms except for the first havehas a factor and thus tend to 0 ash→0. (The first term does not depend onh.) We find that

f(x)=Cn,1xⁿ⁻¹ WithC_n,1=n, this is then

f(x)=nxⁿ⁻¹ which proves the power rule.

EXAMPLE 1 We apply the power rule to various functions and take the opportunity to practice the different notations.

(a) If f(x)=x⁶, then f(x)=⁶x⁵. (b) If f(x)=x³⁰⁰, then f(x)=³⁰⁰x²⁹⁹. (c) Ifg(t)=t⁵, then_dt^dg(t)=⁵t⁴. (d) Ifz=s³, then^{d z}_ds =³s². (e) Ifx = y⁴, then ^{d x}_{d y} =4y³.

Example 1 illustrates the importance of knowing how the variables depend on

y f(x)

x y

Dependent variable

Independent variable

dy

Slopedx

Figure 4.21 Ify= f(x)^{, then}xis the independent variable andyis the dependent variable.

each other (Figure 4.21). Ify = f(x)^{, we call}x theindependent variableandythe dependent variable, becauseydepends on the variablex. For instance, in (a),yis a function ofx; thus,xis the independent, andyis the dependent, variable. In (e), by contrast,xis a function of y; thus, yis now the independent, andxthe dependent, variable. The Leibniz notation ^{d y}_{d x} emphasizes this dependence. When we write ^{d y}_{d x}, we consideryto be a function ofx(i.e.,yis the dependent, andxis the independent, variable) and differentiateywith respect tox.

Since polynomials and rational functions are built up by the basic operations of addition, subtraction, multiplication, and division operating on power functions of the formy = xⁿ,n =⁰,¹,², . . ., we need differentiation rules for such operations.

We begin with the following rules:

Theorem Supposeais a constant and f(x)^andg(x)are differentiable atx. Then the following relationships hold:

1. d

d x[a f(x)] =a d d x f(x) 2. d

d x[f(x)+g(x)] = d

d x f(x)+ d d xg(x)

Rule 1 says that a constant factor can be pulled out of the derivative expression;

Rule 2 says that the derivative of a sum of two functions is equal to the sum of the derivatives of the functions. Similarly, since f(x)−g(x) = f(x)+(−¹)g(x)^{, the} derivative of a difference of functions is the difference of the derivatives:

d

d x[f(x)−g(x)] = d

d x[f(x)+(−¹)g(x)] = d

d x f(x)+ d

d x[(−¹)g(x)]

Using Rule 1 on the rightmost term, we find that _{d x}^d [(−¹)g(x)] = (−¹)_{d x}^d g(x)^. Therefore,

d

d x[f(x)−g(x)] = d

d x f(x)− d d xg(x)

Rules 1 and 2 allow us to differentiate polynomials, as illustrated in the next three examples.

EXAMPLE 2 Differentiatey =²x⁴−³x³+x−^7.

Solution

d

d x(²x⁴−³x³+x−⁷)= d

d x(²x⁴)− d

d x(³x³)+ d d xx− d

d x⁷

=² d

d xx⁴−³ d

d xx³+ d d xx− d

d x⁷

=²(⁴x³)−³(³x²)+¹−⁰=⁸x³−⁹x²+¹

EXAMPLE 3 (a) _{d x}^d (−⁵x⁷+²x³−¹⁰)= −³⁵x⁶+⁶x² (b) _dt^d(t³−8t²+3t)=3t²−16t+3

(c) Suppose that n is a positive integer and a is a constant. Then _ds^d (asⁿ) = ansⁿ⁻¹.

(d) _{d N}^d (^{ln 5}+Nln 7)=^{ln 7}

(e) _dr^d(r²sin^π₄ −r³cos₁₂^π +^sinπ₆)=²rsin^π₄ −³r²cos₁₂^π

In the previous section, we related the derivative to the slope of the tangent line;

the next example uses this interpretation.

EXAMPLE 4 Tangent and Normal Lines If f(x)=²x³−³x+1, find the tangent and normal lines at(−¹,²)^.

Solution

The slope of the tangent line at(−¹,²)^is f(−¹). We begin calculating this derivative as follows:

f(x)=6x²−3 Evaluating f(x)atx = −1, we get

f(−1)=6(−1)²−3=3 Therefore, the equation of the tangent line at(−¹,²)^is

y−²=³(x−(−¹)), ^or y=³x+⁵

To find the equation of the normal line, recall that the normal line is perpendicular to the tangent line; hence, the slopemof the normal line is given by

m = − ¹

f(−¹) = −¹ 3

The normal line goes through the point(−¹,²)as well. The equation of the normal line is therefore

y−²= −¹

3(x−(−¹)), ^or y= −¹ 3x+⁵

3

The graph of f(x), including the tangent and normal lines at(−¹,²), is shown in Figure 4.22.

y

1

2 1 3 4

3 x

4 1 1 2 3 4 Normal line Tangent line

2x³ 3x 1 (1, 2)

2

Figure 4.22 The graph of f(x)=2x³−3x+1, together with the tangent and normal lines at(−1,2).

Look again at the last example: When we computed f(−¹)^{, wefirstcomputed} f(x)^{; thesecond}step was to evaluate f(x)^atx = −1. It makes no sense to plug

−^{1 into} f(x)and then differentiate the result. Since f(−¹) = 2 is a constant, the derivative would be 0, which is obviously not f(−¹). Just look at Figure 4.22 to convince yourself. The notation f(−¹)means that we evaluate the function f(x)^at x = −^1.