The Derivatives of Rational Powers of x
Rational powers ofxare basic building blocks for a rich collection of functions. The rule for the dierentiation of powers ofxis easy to remember:
THE POWER RULE If ris a nonzero rational number, then d
dxxr=rxu1 provided thatxu andxu1are dened.
We will prove the power rule if the exponentris a positive or negative integer. You can nd the proof for an arbitrary rational number at the end of this section.
Let’s begin with the case of a positive integer.n. Bythe Binomial Theorem(as reviewed in Section A2 of Appendix A),
(x+h)n=xn+nxn1h+n(n1)
2 xn2h2+ã ã ã+ n
k
ả
xnkhk+ã ã ã+hn. For anyxRandh6= 0,
f(x+h)f(x)
h = (x+h)nxn h
=
xn+nxn1h+n(n1)
2 xn2h2+ã ã ã+hn
ả xn h
= h
nxn1+ n(n1)
2 xn2h+ã ã ã+hn1
ả h
=nxn1+n(n1)
2 xn2h+ã ã ã+hn1.
108 CHAPTER 2. THE DERIVATIVE Therefore,
f0(x) = lim
h0
nxn1+n(n1)
2 xn2h+ã ã ã+hn1
ả
=nxn1.
The case of a negative integer exponent follows from the rst case. Letr=n, wheren is a positive integer. Iff(x) =xr =xn, we need to show that
f0(x) =rxr1=nxn1
for anyx6= 0so thatf(x)is dened. Ifh6= 0and|h|is small enough f(x+h)is also dened.
The relevant dierence quotient is f(x+h)f(x)
h = (x+h)nxn
h = 1
h 1
(x+h)n 1 xn
ả
= 1 h
xn(x+h)n (x+h)nxn
ả
=
(x+h)nxn h
ả 1 (x+h)nxn
ả . Therefore,
f0(x) = lim
h0
f(x+h)f(x)
h = lim
h0
(x+h)nxn h
ả 1 (x+h)nxn
ả
=
lim
h0
(x+h)nxn h
ả
h0lim 1 (x+h)nxn
ả . By the rst case,
lim
h0
(x+h)nxn
h = d
dxxn=nxn1, andlimh0(x+h)n=xn. Therefore,
f0(x) =¡
nxn1¢ 1 xnxn
ả
=nxn1
x2n =nxn1.
¥
Example 1 Letf(x) =x4. Thenf0(x) = 4x3. Figure 1 shows the graphs off andf0. ¤
2 1 1 2 x 5
10 15 y
fxx4
2 1 1 2 x 20
20 y
f'x4x3
Figure 1: f(x) =x4,f0(x) = 4x3
Example 2 Let
f(x) = 1 x2. a) Determinef0.
b) Determine the points at whichf is not dierentiable and the relevant limits off0. Interpret the results graphically.
Solution
a) By the power rule,
f0(x) = d dx
¡x2¢
=2x3= 2 x3. The above expression is valid for eachx6= 0.
b) The functionf is dierentiable at x6= 0. Sincef(x)is not dened atx= 0, f is certainly not dierentiable at 0. On the other hand, we can discuss the limits off0(x)asxapproaches 0 from the right and from the left.
We have
f0(x) = 2
x3 = (2) 1
x3
ả . Sincex3>0ifx >0andlimx0x3= 0,
x0+lim 1
x3 = +. Sincelimx0(2) =2<0,
x0+lim f0(x) = lim
x0+(2) 1
x3
ả
=. On the other hand,x3>0if x >0so that
x0lim 1
x3 = Therefore,
x0lim f0(x) = lim
x0+(2) 1
x3
ả
= +.
Therefore, the vertical axis is a vertical asymptote for the graphs off0(andf). Figure 2 shows the graphs of f and f0. The picture is consistent with our analysis. The tangent line to the graph off at (x, f(x)) becomes steeper and steeper as x approaches 0 from either side. The slope is negative to the right of 0, and positive to the left of 0. This example is a prototype for 1/xnwherenis an even positive integer. ¤
2 1 1 2 5
10 15
fx1 x2
2 1 1 2
20 10 10 20
f'x 2 x3
Figure 2
110 CHAPTER 2. THE DERIVATIVE
Example 3 Letf(x) =
x=x1/2.
a) Determinef0 directly from the denition of the derivative.
b) Show thatf is not dierentiable at 0 from the right.
c) Determinelimx0+f0(x). Interpret the result graphically and in terms of the rate of change off.
d) Determinelimx+f0(x). Interpret the result as in part c) Solution
a) We will use a time-honored trick to express the relevant dierence quotient in a way that will lead to the derivative off:
f(x+h)f(x)
h =
x+h x h
=
x+h x h
ả
x+h+ x
x+h+ x
ả
= (x+h)x h¡
x+h+ x¢
= h
h¡
x+h+ x¢
= 1
x+h+ x. Therefore,
f0(x) = lim
h0
f(x+h)f(x)
h = lim
h0
1
x+h+ x = 1
2 x for anyx >0.
b) We have
h0+lim
f(0 +h)f(0)
h = lim
h0+
h h = lim
h0+
1
h = +.
Therefore,f is not dierentiable at 0 from the right even though it is continuous at 0 from the right.
c) We also have
x0+lim f0(x) = lim
x0+
1 2
x = +.
The tangent line to the graph off at(x, f(x))becomes steeper and steeper asxapproaches 0 from the right. Sincef0(x)is the rate of change off atx, the rate of change of the square-root functionf increases beyond all bounds asxapproaches 0 from the right.
d) We have
x+lim f0(x) = lim
x+
1 2
x = 0.
Thus, the slope of the tangent line to the graph of the square-root function at(x,x)approaches 0 asxbecomes larger and larger. Sincef0(x)is the rate of change off atx, the rate of change of the square root function decreases towards to 0 asxbecomes large.
Figure 3 shows the graphs of f and f0. The picture is consistent with our analysis. Since limx0+f0(x) = +, the vertical axis is a vertical asymptote for the graph off0. The square- root function is a prototype of functions dened byx1/n, wherenis an even positive integer.
¤
4 8 x
1 2 y
fx x
4 8 x
1 2 y
f'x 1
2 x
Figure 3: f(x) =
x, f0(x) = 1 2 x
Example 4 Letf(x) =x1/3. a) Determinef0.
b) Show thatf is not dierentiable at 0.
c) Determinelimx0f0(x). Interpret the result graphically.
Solution
a) By the power rule,
f0(x) = d dx
³ x1/3
´= 1
3x1/31= 1
3x2/3= 1 3x2/3. The above expression is valid ifx6= 0.
b) We have
h0lim
f(0 +h)f(0)
h = lim
h0
h1/3 h = lim
h0
1
h2/3 = +, since h2/3 = ¡
h1/3¢2
> 0 for each h 6= 0 and limh0h2/3 = 0. Thus, the function is not dierentiable at 0, even though it is continuous at 0.
c) We also have
x0limf0(x) = lim
x0
1
3x2/3 = +.
Therefore, the tangent line to the graph of f at (x, f(x)) becomes steeper and steeper asx approaches 0. The slope is positive on either side of 0. Sincelimx0f0(x) = +, the vertical axis is a vertical asymptote for the graph of f. Figure 4 shows the graphs of f and f0. The picture is consistent with our analysis. The cube-root function is a prototype of functions dened byx1/n, wherenis an odd positive integer. ¤
112 CHAPTER 2. THE DERIVATIVE
4 8 x
2 1 1 2 y
fxx13
8 4 4 8x 0.4
0.2 y
f'x 1 3x23
Figure 4: f(x) =x1/3, f0(x) = 1 3x2/3
There is terminology that describes the behavior of the derivative of a function that is similar to the behavior of the derivatives of the square-root function and the cube-root function near 0:
Denition 1 The graph off has avertical tangent at(a, f(a))if f is continuous ataand lim
xaf0(x) =±, or
f(x) is dened only ifxa,f is continuous atafrom the right, and lim
xa+f0(x) =±, or
f(x)is dened only ifx a, f is continuous atafrom the left, and lim
xaf0(x) =±. Example 5 Show that the graph off has a vertical tangent at 0 if
a) f(x) =
x,b)f(x) =x1/3. Solution
a) Letf(x) =
x. Thenf(x)is dened only forx0andf is continuous at 0 from the right.
In Example 3 we showed that limx0+f0(x) = +. Therefore, the graph off has a vertical tangent at 0. Indeed, in Figure 3 the vertical axis appears to be tangential to the graph off at (0,0) = (0, f(0)).
b) Let f(x) = x1/3. Then f(x) is dened for each x R and f is continuous at 0. In Example 4 we showed that limx0f(x) = +. Therefore, the graph of f has a vertical tangent at 0. Indeed, in Figure 4 the vertical axis appears to be tangential to the graph off at (0,0) = (0, f(0)). ¤
Example 6 Letf(x) =x2/3. a) Determinef0.
b) Show thatf is not dierentiable at 0.
c) Determinelimx0+f0(x)and limx0f0(x). Interpret the result graphically.
Solution
a) By the power rule,
f0(x) = d
dxx2/3= 2
3x1/3= 2 3x1/3. The above expression is valid ifx6= 0.
b) We have
h0+lim
f(0 +h)f(0)
h = lim
h0+
h2/3 h = lim
h0+
1
h1/3 = +,
sinceh1/3>0if h >0and limh0h1/3= 0. Therefore, the function is not dierentiable at 0.
c) Similarly,
x0+lim f0(x) = lim
x0+
2
3x1/3 = +. We have
x0lim f0(x) = lim
x0
2
3x1/3 =, sincex1/3<0if x <0andlimx0x1/3= 0.
Therefore, the tangent line to the graph of f at (x, f(x)) becomes steeper and steeper asx approaches 0. The sign of the slope is positive to the right of the origin and negative to the left of the origin.
Figure 5 shows the graphs off andf0. Sincelimx0+f0(x) = +andlimx0f0(x) =, the vertical axis is a vertical asymptote for the graph off0. ¤
8 4 4 8 x 2
4 y
fxx23
8 4 4 8 x 1
1 y
f'x 2 3x13
Figure 5: f(x) =x2/3, f0(x) = 2 3x1/3
There is terminology that describes the behavior of the derivative of a function that is similar to the behavior of the derivative of the function in Example 6:
Denition 2 The graph off has acuspat(a, f(a))iff is continuous ata, and
xa+lim f0(x) = +, lim
xaf0(x) =, or
xa+lim f0(x) =and lim
xaf0(x) = +.
Example 7 Letf(x) =x2/3, as in Example 6. Sincelimx0+f0(x) = +andlimx0f0(x) = , the graph off has a cusp at(0,0) = (0, f(0)). Figure 5 is typical when the graph of a function has a cusp. If a function f has a cusp at (a, f(a)), the graph of f seems to have a
“sharp beak” at(a, f(a)). ¤
114 CHAPTER 2. THE DERIVATIVE In the expression
d
dxxr =xr1,
the derivative should be interpreted as a right derivative atx= 0ifxr and xr1are dened if and only ifx0. For example, iff(x) =x5/4=¡
x1/4¢5
, then the domain off is the interval [0,+). By the power rule,
f0(x) = d dx
³ x5/4
´= 5 4x1/4.
The above expression denesf0(x)ifx >0. Both expressionx5/4andx1/4are dened atx= 0 and have the value 0. We havef+0 (0) = 0. Figure 6 shows the graphs of f andf0.
2 4 6 8 x
4 8 12 y
fxx54
2 4 6 8 x
1 2 y
f'x5 4x14
Figure 6: f(x) =x5/4, f0(x) = 5 4x1/4
The Derivatives of Linear Combinations
THE CONSTANT MULTIPLE RULE FOR DIFFERENTIATION Assume that f is dierentiable at x, and thatc is a constant. Then, cf is also dierentiable at x, and we have
(cf)0(x) =cf0(x).
In the Leibniz notation,
d
dx(cf(x))=c d dxf(x).
Proof
The dierence quotient corresponding tocf,xandh6= 0is (cf) (x+h)(cf) (x)
h = cf(x+h)cf(x)
h =c
f(x+h)f(x) h
ả . By the constant multiple rule for limits,
(cf)0(x) = lim
h0
c
f(x+h)f(x) h
ảả
=clim
h0
f(x+h)f(x)
h =cf0(x).
¥
Example 8 Letf(x) =x3. Then, d
dx(2f(x)) = d dx
¡2x3¢= 2 d
dx
âx3đả= 2â3x2đ= 6x.
¤
The derivative of a sum of functions is the sum of their derivatives:
THE SUM RULE FOR DIFFERENTIATION Assume thatf andg are dieren- tiable atx. Then, the sumf+g is also dierentiable at x, and we have
(f+g)0(x) =f0(x) +g0(x).
In the Leibniz notation, d
dx (f(x) +g(x)) = d
dxf(x)+ d dxg(x).
Proof
The dierence quotient corresponding tof+g,xand h6= 0is (f+g) (x+h)(f+g) (x)
h = f(x+h) +g(x+h)(f(x) +g(x)) h
= f(x+h) +g(x+h)f(x)g(x) h
= f(x+h)f(x)
h +g(x+h)g(x)
h .
By the sum rule for limits,
(f+g)0(x) = lim
h0
f(x+h)f(x)
h +g(x+h)g(x) h
ả
= lim
h0
f(x+h)f(x)
h + lim
h0
g(x+h)g(x) h
=f0(x) +g0(x).
¥
Example 9
d dx
¡x+x2¢
= d
dx(x) + d dx
¡x2¢
= 1 + 2x, by the sum rule and the power rule. ¤
The sum rule extends to the sum of an arbitrary number of functions. Thus, d
dx(f1(x) +f2(x) +ã ã ã+fn(x)) = d
dxf1(x) + d
dxf2(x) +ã ã ã+ d dxfn(x), wherenis an arbitrary positive integer.
Recall that alinear combinationof the functionsf andgis a function of the formc1f+c1g, where c1 and c2 are constants. The derivative of a linear combination of functions is the linear combination of the corresponding derivatives, with the same coecients:
116 CHAPTER 2. THE DERIVATIVE DIFFERENTIATION IS A LINEAR OPERATION Assume that f and g are dierentiable atx. If c1andc2are constants, then the linear combinationc1f+c2g is also dierentiable at x, and we have
(c1f +c2g)0(x) =c1f0(x) +c2g0(x).
In the Leibniz notation, d
dx(c1f(x) +c2g(x)) =c1 d
dxf(x) +c2 d dxg(x).
Proof
We apply the sum rule and the constant multiple rule for dierentiation:
d
dx(c1f(x) +c2g(x)) = d
dx(c1f(x)) + d
dx(c2g(x)) =c1 d
dxf(x) +c2 d dxg(x).
¥
The above rule extends to linear combinations of any number of functions:
d
dx(c1f1x) +c2f2(x) +ã ã ã+cnfn(x)) =c1 d
dxf1(x) +c2 d
dxf2(x) +ã ã ã+cn d dxfn(x), wherec1, c2, . . . , cn are constants.
Example 10 Let
f(x) = 11 2x2+ 1
24x4. a) Determinef0(x).
b) Determine the tangent line to the graph off at(1, f(1)).
Solution
a) By the linearity of dierentiation and the power rule, d
dxf(x) = d
dx(1)1 2
d dx
¡x2¢ + 1
24 d dx
¡x4¢
= 01
2(2x) + 1 24
¡4x3¢
=x+ 1 6x3. b) We have
f(1) = 13
24 andf0(1) =23 24
Therefore, the tangent line to the graph off at(1, f(1))is the graph of the equation y=f(1) +f0(1) (x1) = 13
2423
24(x1).
Figure 7 shows the graph off and the tangent line to the graph off at(1, f(1)). ¤
4 2 1 2 4 x 2
4 6 8 y
1,f1
Figure 7
A polynomial is a linear combination of the constant 1 and positive integer powers of x.
Therefore, we can dierentiate any polynomial, as in Example 10, thanks to the linearity of dierentiation: If
f(x) =c0+c1x+c2x2+ã ã ã+cnxn, wherec0, c1, . . . , cn are given constants, then
f0(x) =c1+ 2c2x+ã ã ã+ncnxn1. Note thatf0(x)is a polynomial of degree n1.
Example 11 Let
f(x) = 3x2+ 1 x Determinef0(x).
Solution
By the linearity of dierentiation and the power rule, df
dx = d dx
³3x2+x1/2
´= 3d dx
¡x2¢ + d
dx
³ x1/2
´
= 3 (2x) 1
2x3/2= 6x 1 2x3/2. Note that the expression is valid ifx >0.¤
Higher-Order Derivatives
The second derivative of a functionf is the derivative off0. In the “prime notation”, the second derivative is denoted asf00. Thus,f00(x) = (f0)0(x)iff0 is dierentiable atx. If we use the Leibniz notation to denote the derivative, we have
f00(x) = d dx
d dxf(x)
ả . This suggests the Leibniz notation
d2f dx2 for the second derivative off. We may type this as
d2f
dx2(x), d2f(x) dx2 or d2
dx2f(x).
118 CHAPTER 2. THE DERIVATIVE Just as in the case of the derivative, the Leibniz notation for the second derivative is convenient to use, as long as you don’t try to attach a meaning other than
d dx
d dxf(x)
ả
to the symbol
d2f dx2.
The above expression doesnot involve raising a quantityd/dxto the second power.
The derivative off00isthe third derivativef000off: f000(x) = (f00)0(x). The Leibniz notation for the third derivative is
d3f dx3. Thus,
d3f dx3 = d
dx d2
dx2f(x)
ả . This may be typed as
d3f(x) dx3 or d3
dx3f(x).
The second derivative of f is also referred to as the second-order derivative of f, and the third derivative is the third-order derivative of f. More generally, we obtainthe nth order derivative of f by dierentiating the derivative of order n1. As n increases, the prime notation becomes unwieldy. We may denote thenth order derivative off byf(n). There is no diculty to express higher-order derivatives by the Leibniz notation:
f(n)(x) = dn
dxnf(x) = d dx
dn1 dxn1f(x)
ả . Example 12 Letf(x) =x4. Determine the second derivative off.
Solution
By the power rule,
f0(x) = d dx
¡x4¢= 4x3,
f00(x) =d2f
dx2(x) = d dx
d dxf(x)
ả
= d dx
¡4x3¢= 4d dx
¡x3¢= 4¡3x2¢= 12x2,
f(3)(x) = d3f
dx3(x) = d dx
d2 dx2f(x)
ả
= d dx
¡12x2¢
= 12d dx
¡x2¢
= 12 (2x) = 24x, f(4)(x) = d4f
dx4(x) = d dx
d3 dx3f(x)
ả
= d
dx(24x) = 24, and
f(n)(x) = dnf
dxn(x) = 0 forn= 5,6,7, . . . ¤
The Proof of the Power Rule for Arbitrary Rational Powers
We have already established the power rule for positive and negative integer powers ofx. Let’s begin by deriving the rule for the derivative of x1/n for any positive integern. We will only considerx >0. The proof is similar ifx <0 andx1/nis dened.
The dierence quotient that is relevant to the calculation of d
dxx1/n is
(x+ x)1/nx1/n
x ,
wherex6= 0.
x y
y=x1 n y+Dy=Hx+DxL1 n
x Dxx+Dx Dy
Figure 8 With reference to Figure 8, let’s set
y=x1/nand y+ y= (x+ x)1/n, so that
x=yn, x+ x= (y+ y)n.andx= (y+ y)nyn. Thus,x= (y+ y)nyn. Note thaty6= 0sincex6= 0. Therefore,
(x+ x)1/nx1/n
x = (y+ y)y
(y+ y)nyn = y
(y+ y)nyn = 1 (y+ y)nyn
y By the continuity of the function dened byx1/n,
x0lim y= lim
x0
³(x+ x)1/nx1/n
´= 0.
Therefore,
x0lim
(x+ x)1/nx1/n
x = lim
x0
1 (y+ y)nyn
y
= 1
limx0(y+ y)nyn y
= 1
limy0(y+ y)nyn y
,
120 CHAPTER 2. THE DERIVATIVE provided that
y0lim
(y+ y)nyn
y 6= 0.
By the power rule,
y0lim
(y+ y)nyn
y = d
dy(yn) =nyn1>0 sincey=x1/n>0. Therefore,
d dx
³ x1/n´
= lim
x0
(x+ x)1/nx1/n
x = 1
nyn1 = 1
n¡
x1/n¢n1 = 1
nx11/n = 1 nx1/n1 ifx >0. ¥
Now we are ready to establish the rule for an arbitrary rational number. Thus, let r= m/n wherenis a positive integer andma positive or negative integer. The dierence quotient that is relevant to the calculation of
d dxxm/n
is (x+ x)m/nxm/n
x ,
wherex6= 0. Let’s setu=x1/n and(x+ x)1/n=u+ u.
(x+ x)m/nxm/n
x = (u+ u)mum
x
=
(u+ u)mum u
ả u x
ả . Sinceu= (x+ x)1/nx1/n, and x1/ndenes a continuous function,
x0lim u= 0.
Therefore,
x0lim
(x+ x)m/nxm/n
x =
x0lim
(u+ u)mum u
ả
x0lim u x
ả
=
u0lim
(u+ u)mum u
ả du dx
ả
= d
duum
ả d dx
³ x1/n
´ả
= ¡
mum1¢ 1 nx1/n1
ả
= m
n
³
x1/n´m1³
x1/n1´
= m
nxm/n1/n+1/n1
= m
nxm/n1=rxr1.
¥
Problems
In problems 1-6, determine the indicated derivative by using the power rule. You need to specify the domain of the derivative function.
1. d dx
¡x5¢
2. d dx
1 x3
ả
3. d dx
¡x1/7¢
4. d dx
¡x5/4¢
5. d dx
¡x3/5¢
6. d dx
1 x1/6
ả
In problems 7-12,
a) Determinef0. You need to specify the domain of the derivative function.
b) Determine the points at which the graph off has a vertical tangent or a cusp. You need to justify your assertions.
c) [C] Plot the graphs of f and f0 with the help of your graphing utility. Are the pictures consistent with your assertions in parts a) and b)?
7. f(x) =x33 8. f(x) = 1
x+ 4 9. f(x) =x1/4+ 2
10. f(x) =x1/72 11. f(x) =x3/4 12. f(x) =x4/5+ 1 In problems 13 and 14,
a) Determine the tangent line to the graph off at(a, f(a))(the point-slope form with basepoint awill do).
b)[C]Make use of your graphing utility to plot the graph off and the tangent line at(a, f(a)).
Zoom in towards(a, f(a))until you are unable to distinguish between the graph off and the tangent line. Do the pictures reinforce the identication of the slope of the graph offat(a, f(a)) with the slope of the tangent line at that point?
13. f(x) =x2/3, a= 8 14. f(x) =x1/4, a= 16 In problems 15 - 20, determinef0. You need not specify the domain off0. 15. f(x) =4
x 16. f(x) = 2
x1/3
17. f(x) = 3x24x+ 8
18. f(x) = 5x47x2+ 3x10 19. f(x) = 2x3
x+ 9x2/3 20. f(x) = 4
x7/410x5+ 100