7001_AWLThomas_ch03p122-221.qxd 10/12/09 2:21 PM Page 122
DIFFERENTIATION
Tangents and the Derivative at a Point
y
Finding a Tangent to the Graph of a Function
y f (x)
Q(x 0 h, f (x 0 h))
f (x 0 h) f (x 0)
To find a tangent to an arbitrary curve y = ƒ(x) at a point Psx0 , ƒsx0 dd , . We calculate the
slope of the secant through P and a nearby point Qsx0 + h, ƒsx0 + hdd.
We then investigate the limit of the slope as h : 0 (Figure 1). If the limit exists, we
call it the slope of the curve at P and define the tangent at P to be the line
through P having this slope.
P(x 0, f(x 0))
h
x
x0 h
0
DEFINITIONS
The slope of the curve y = ƒsxd at the point Psx0 , ƒsx0 dd is the
number
x0
ƒsx0 + hd - ƒsx0 d
h
h:0
FIGURE 1
m = lim
(provided the limit exists).
The tangent line to the curve at P is the line through P with this slope.
y
y 5 1x
EXAMPLE
slope is – 12
a
0
a
x
(a) Find the slope of the curve y = 1>x at any point x = a Z 0. What is the slope at the
point x = - 1?
(b) Where does the slope equal - 1>4?
(c) What happens to the tangent to the curve at the point (a, 1>a) as a changes?
Solution
slope is –1
at x 5 –1
FIGURE 2 The tangent slopes, steep
near the origin, become more gradual as
the point of tangency moves away
(Example 1).
(a) Here ƒsxd = 1>x. The slope at (a, 1>a) is
1
1
- a
ƒsa + hd - ƒsad
a + h
1 a - sa + hd
= lim
= lim
lim
h
h
h:0
h:0
h:0 h asa + hd
= lim
h:0
-h
-1
1
= lim
= - 2.
hasa + hd
asa
+
hd
h:0
a
7001_AWLThomas_ch03p122-221.qxd 10/12/09 2:21 PM Page 123
Notice how we had to keep writing “limh:0” before each fraction until the stage
where we could evaluate the limit by substituting h = 0. The number a may be positive or negative, but not 0. When a = - 1, the slope is - 1>(-1) 2 = - 1 (Figure 2).
(b) The slope of y = 1>x at the point where x = a is -1>a 2. It will be -1>4 provided
that
y
y 1x
⎝
⎛2, 1 ⎛
⎝ 2
slope is – 1
4
x
slope is – 1
4
⎛–2, – 1 ⎛
⎝
2⎝
1
1
= - .
4
a2
This equation is equivalent to a 2 = 4, so a = 2 or a = - 2. The curve has slope
- 1>4 at the two points (2, 1>2) and s - 2, -1>2d (Figure 3).
(c) The slope -1>a 2 is always negative if a Z 0. As a : 0 +, the slope approaches - q
and the tangent becomes increasingly steep (Figure 2). We see this situation again as
a : 0 - . As a moves away from the origin in either direction, the slope approaches 0
and the tangent levels off to become horizontal.
FIGURE 3 The two tangent lines to
y = 1>x having slope -1>4 (Example 1).
The Derivative as a Function
y f(x)
Secant slope is
f (z) f (x)
zx
Q(z, f(z))
ƒ¿sxd = lim
h:0
ƒsx + hd - ƒsxd
,
h
provided the limit exists.
f (z) f (x)
P(x, f(x))
hzx
x
DEFINITION The derivative of the function ƒ(x) with respect to the variable x is
the function ƒ¿ whose value at x is
zxh
Derivative of f at x is
f (x h) f(x)
f '(x) lim
h
h→0
lim f(z) f (x)
z→x
zx
FIGURE 4 Two forms for the difference
quotient.
If we write z = x + h, then h = z - x and h approaches 0 if and only if z approaches x.
Therefore, an equivalent definition of the derivative is as follows (see Figure 4). This
formula is sometimes more convenient to use when finding a derivative function.
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EXAMPLE 1
Differentiate ƒsxd =
x
.
x - 1
solution We use the definition of derivative, which requires us to calculate
ƒ(x + h) and then subtract ƒ(x) to obtain the numerator in the difference quotient.
ƒsxd =
We have
x
x - 1
and
ƒsx + hd =
sx + hd
, so
sx + hd - 1
ƒsx + hd - ƒsxd
h
h:0
ƒ¿sxd = lim
Definition
x
x + h
x - 1
x + h - 1
= lim
h
h:0
= lim
1 # sx + hdsx - 1d - xsx + h - 1d
h
sx + h - 1dsx - 1d
c
ad - cb
a
- =
b
d
bd
= lim
-h
1 #
h sx + h - 1dsx - 1d
Simplify.
h:0
h:0
-1
-1
.
=
h:0 sx + h - 1dsx - 1d
sx - 1d2
= lim
Cancel h. Z 0
EXAMPLE 2
(a) Find the derivative of ƒsxd = 1x for x 7 0.
(b) Find the tangent line to the curve y = 1x at x = 4.
Solution
y
(a) We use the alternative formula to calculate ƒ¿ :
y 1x1
4
ƒszd - ƒsxd
z - x
z :x
ƒ¿sxd = lim
(4, 2)
y 兹x
1
0
= lim
4
x
z :x
= lim
FIGURE 5 The curve y = 1x and its
tangent at (4, 2). The tangent’s slope is
found by evaluating the derivative at x = 4
(Example 2).
z :x
= lim
z :x
1z - 1x
z - x
1z - 1x
A 1z - 1x B A 1z + 1x B
1
1
=
.
1z + 1x
21x
(b) The slope of the curve at x = 4 is
ƒ¿s4d =
1
.
4
The tangent is the line through the point (4, 2) with slope 1>4 (Figure 5):
1
y = 2 + sx - 4d
4
1
y = x + 1.
4
22
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Differentiation Rules
y
c
(x h, c)
(x, c)
Derivative of a Constant Function
If ƒ has the constant value ƒsxd = c, then
yc
dƒ
d
=
scd = 0.
dx
dx
h
0
x
xh
FIGURE 6 The rule sd>dxdscd = 0 is
another way to say that the values of
constant functions never change and that
the slope of a horizontal line is zero at
every point.
x
Proof We apply the definition of the derivative to ƒsxd = c, the function whose outputs
have the constant value c (Figure 6). At every value of x, we find that
ƒ¿sxd = lim
h:0
ƒsx + hd - ƒsxd
c - c
= lim
= lim 0 = 0.
h
h
h:0
h:0
Power Rule (General Version)
If n is any real number, then
d n
x = nx n - 1,
dx
for all x where the powers x n and x n - 1 are defined.
EXAMPLE 1
(a) x 3
Differentiate the following powers of x.
(b) x 2/3
(c) x 22
(d)
1
x4
(e) x -4>3
(f) 2x 2 + p
Solution
(a)
d 3
(x ) = 3x 3 - 1 = 3x 2
dx
(b)
d 2>3
2
2
(x ) = x (2>3) - 1 = x -1>3
3
3
dx
(c)
d 22
A x B = 22x 22 - 1
dx
(d)
d 1
d -4
4
a b =
(x ) = - 4x -4 - 1 = - 4x -5 = - 5
dx x 4
dx
x
(e)
d -4>3
4
4
(x
) = - x -(4>3) - 1 = - x -7>3
3
3
dx
(f) d 2x 2 + p = d x 1 + (p>2) = a1 + p bx 1 + (p>2) - 1 = 1 (2 + p) 2x p
A
B dx A
B
2
2
dx
Derivative Constant Multiple Rule
If u is a differentiable function of x, and c is a constant, then
du
d
scud = c .
dx
dx
In particular, if n is any real number, then
d
scx n d = cnx n - 1 .
dx
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EXAMPLE 2
(a) The derivative formula
d
s3x 2 d = 3 # 2x = 6x
dx
Derivative Sum Rule
If u and y are differentiable functions of x, then their sum u + y is differentiable
at every point where u and y are both differentiable. At such points,
d
du
dy
su + yd =
+
.
dx
dx
dx
For example, if y = x 4 + 12x, then y is the sum of u(x) = x 4 and y(x) = 12x. We
then have
dy
d
d
=
(x 4) +
(12x) = 4x 3 + 12.
dx
dx
dx
EXAMPLE 3
Solution
Find the derivative of the polynomial y = x 3 +
dy
d 3
d 4 2
d
d
=
x +
a x b s5xd +
s1d
dx
dx
dx 3
dx
dx
= 3x 2 +
4 2
x - 5x + 1.
3
Sum and Difference Rules
8
4#
2x - 5 + 0 = 3x 2 + x - 5
3
3
EXAMPLE 4
Does the curve y = x 4 - 2x 2 + 2 have any horizontal tangents? If so,
where?
Solution
y
y x 4 2x 2 2
dy
d 4
=
sx - 2x 2 + 2d = 4x 3 - 4x .
dx
dx
dy
= 0 for x:
Now solve the equation
dx
(0, 2)
(–1, 1)
–1
1
0
4x 3 - 4x = 0
4xsx 2 - 1d = 0
x = 0, 1, -1.
(1, 1)
1
The horizontal tangents, if any, occur where the slope dy>dx is zero. We have
x
FIGURE 7 The curve in Example 4
and its horizontal tangents.
The curve y = x 4 - 2x 2 + 2 has horizontal tangents at x = 0, 1, and -1. The corresponding points on the curve are (0, 2), (1, 1) and s -1, 1d . See Figure 7.
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Derivative Product Rule
If u and y are differentiable at x, then so is their product uy, and
d
dy
du
+ y .
suyd = u
dx
dx
dx
The derivative of the product uy is u times the derivative of y plus y times the derivative of u. In prime notation, suyd¿ = uy¿ + yu¿ . In function notation,
d
[ƒsxdgsxd] = ƒsxdg¿sxd + gsxdƒ¿sxd.
dx
EXAMPLE 6
Find the derivative of y = sx 2 + 1dsx 3 + 3d .
Solution
(a) From the Product Rule with u = x 2 + 1 and y = x 3 + 3, we find
d
C sx 2 + 1dsx 3 + 3d D = sx 2 + 1ds3x 2 d + sx 3 + 3ds2xd
dx
d
dy
du
+ y
suyd = u
dx
dx
dx
= 3x 4 + 3x 2 + 2x 4 + 6x
= 5x 4 + 3x 2 + 6x.
(b) This particular product can be differentiated as well by multiplying out the
original expression for y and differentiating the resulting polynomial:
y = sx 2 + 1dsx 3 + 3d = x 5 + x 3 + 3x 2 + 3
dy
= 5x 4 + 3x 2 + 6x.
dx
Derivative Quotient Rule
If u and y are differentiable at x and if ysxd Z 0, then the quotient u>y is differentiable at x, and
d u
a b =
dx y
y
du
dy
- u
dx
dx
.
2
y
In function notation,
gsxdƒ¿sxd - ƒsxdg¿sxd
d ƒsxd
c
.
d =
dx gsxd
g 2sxd
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EXAMPLE 7
Find the derivative of (a) y =
t2 - 1
,
t3 + 1
y = e -x
Solution
We apply the Quotient Rule with
u = t 2 - 1 and y = t 3 + 1 :
dy
st 3 + 1d # 2t - st 2 - 1d # 3t 2
=
dt
st 3 + 1d2
EXAMPLE 8
=
2t 4 + 2t - 3t 4 + 3t 2
st 3 + 1d2
=
- t 4 + 3t 2 + 2t
.
st 3 + 1d2
ysdu>dtd - usdy>dtd
d u
a b =
dt y
y2
Rather than using the Quotient Rule to find the derivative of
y =
sx - 1dsx 2 - 2xd
,
x4
expand the numerator and divide by x 4 :
y =
sx - 1dsx 2 - 2xd
x 3 - 3x 2 + 2x
=
= x -1 - 3x -2 + 2x -3 .
x4
x4
Then use the Sum and Power Rules:
dy
= - x -2 - 3s - 2dx -3 + 2s - 3dx -4
dx
6
6
1
= - 2 + 3 - 4.
x
x
x
Second- and Higher-Order Derivatives
If y = ƒsxd is a differentiable function, then its derivative ƒ¿sxd is also a function. If ƒ¿ is
also differentiable, then we can differentiate ƒ¿ to get a new function of x denoted by ƒ–.
So ƒ– = sƒ¿ d¿ . The function ƒ– is called the second derivative of ƒ because it is the derivative of the first derivative. It is written in several ways:
ƒ–sxd =
y snd =
d 2y
dx
2
=
dy¿
dy
d
b =
= y–
a
dx dx
dx
d ny
d
y sn - 1d =
= D ny
dx
dx n
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EXAMPLE 10
The first four derivatives of y = x 3 - 3x 2 + 2 are
First derivative:
y¿ = 3x 2 - 6x
Second derivative:
y– = 6x - 6
Third derivative:
y‡ = 6
Fourth derivative:
y s4d = 0 .
The Chain Rule
THEOREM —The Chain Rule
If ƒ(u) is differentiable at the point u = g sxd
and g (x) is differentiable at x, then the composite function sƒ gdsxd = ƒsg sxdd
is differentiable at x, and
sƒ gd¿sxd = ƒ¿sg sxdd # g¿sxd .
In Leibniz’s notation, if y = ƒsud and u = g sxd , then
dy
dy
=
dx
du
# du ,
dx
where dy>du is evaluated at u = g sxd .
EXAMPLE1
The functions y = u2
and u = 3x 2 + 1 Calculate
dy
dx
is the composite of y = ƒ(u) = u 2 and u = g(x) = 3x 2 + 1 . Calculating derivatives, we
see that
dy
=
dx
dy
du
# du
dx
= 2u # 6x
= 2s3x 2 + 1d # 6x
= 36x 3 + 12x .
Calculating the derivative from the expanded formula (3x 2 + 1) 2 = 9x 4 + 6x 2 + 1 gives
the same result:
dy
d
=
(9x 4 + 6x 2 + 1)
dx
dx
= 36x 3 + 12x .
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The Chain Rule with Powers of a Function
If ƒ is a differentiable function of u and if u is a differentiable function of x, then substituting y = ƒsud into the Chain Rule formula
dy
dy du
#
=
dx
du dx
leads to the formula
d
du
.
ƒsud = ƒ¿sud
dx
dx
If n is any real number and ƒ is a power function, ƒsud = u n , the Power Rule tells us
that ƒ¿sud = nu n - 1 . If u is a differentiable function of x, then we can use the Chain Rule to
extend this to the Power Chain Rule:
d
A u n B = nu n - 1
du
d
du
.
su n d = nu n - 1
dx
dx
EXAMPLE 2
The Power Chain Rule simplifies computing the derivative of a power of
an expression.
(a)
d
d
s5x 3 - x 4 d7 = 7s5x 3 - x 4 d6
(5x 3 - x 4)
dx
dx
Power Chain Rule with
u = 5x 3 - x 4, n = 7
= 7s5x 3 - x 4 d6s5 # 3x 2 - 4x 3 d
= 7s5x 3 - x 4 d6s15x 2 - 4x 3 d
(b)
d
d
1
b =
s3x - 2d-1
a
dx 3x - 2
dx
d
s3x - 2d
dx
= - 1s3x - 2d-2s3d
3
= s3x - 2d2
= - 1s3x - 2d-2
Power Chain Rule with
u = 3x - 2, n = - 1
EXAMPLE 3
I we saw that the absolute value function y = ƒ x ƒ is not
differentiable at x 0. However, the function is differentiable at all other real numbers as
we now show. Since ƒ x ƒ = 2x 2, we can derive the following formula:
d
d
2x 2
( x ) =
dx ƒ ƒ
dx
1
# d (x 2)
=
2 dx
2 2x
Power Chain Rule with
u = x 2, n = 1>2, x Z 0
2x 2 = ƒ x ƒ
=
1
2ƒxƒ
# 2x
=
x
,
x
ƒ ƒ
x Z 0.
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EXAMPLE 4
Show that the slope of every line tangent to the curve y = 1>s1 - 2xd3 is
positive.
We find the derivative:
Solution
dy
d
=
s1 - 2xd-3
dx
dx
# d s1
dx
2xd-4 # s - 2d
= - 3s1 - 2xd-4
= - 3s1 =
- 2xd
Power Chain Rule with u = s1 - 2xd, n = - 3
6
.
s1 - 2xd4
At any point (x, y) on the curve, x Z 1>2 and the slope of the tangent line is
dy
6
=
,
dx
s1 - 2xd4
Implicit Differentiation
Implicit Differentiation
1. Differentiate both sides of the equation with respect to x, treating y as a
differentiable function of x.
2. Collect the terms with dy>dx on one side of the equation and solve for dy>dx.
Find dy>dx if y 2 = x .
EXAMPLE 1
The equation y 2 = x defines two differentiable functions of x that we can actually find, namely y1 = 1x and y2 = - 1x (Figure 8). We know how to calculate the
derivative of each of these for x 7 0 :
Solution
y
y2 x
dy1
1
=
dx
2 1x
Slope 1 1
2y1 2x
and
dy2
1
= .
dx
2 1x
y1 x
P(x, x )
x
0
Q(x, x )
y 2 x
Slope 1 1
2y 2
2x
2
FIGURE 8 The equation or y - x = 0 ,
y 2 = x as it is usually written, defines
two differentiable functions of x on the
interval x 7 0. Example 1 shows how to
find the derivatives of these functions
without solving the equation y 2 = x for y.
But suppose that we knew only that the equation y 2 = x defined y as one or more differentiable functions of x for x 7 0 without knowing exactly what these functions were.
Could we still find dy>dx?
The answer is yes. To find dy>dx, we simply differentiate both sides of the equation
y 2 = x with respect to x, treating y = ƒsxd as a differentiable function of x:
y2 = x
dy
= 1
2y
dx
dy
1
=
.
2y
dx
The Chain Rule gives
d
Ay2B =
dx
dy
d
2
.
[ƒsxd] = 2ƒsxdƒ¿sxd = 2y
dx
dx
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This one formula gives the derivatives we calculated for both explicit solutions y1 = 1x
and y2 = - 1x :
dy1
1
1
=
=
2y1
dx
2 1x
dy2
1
1
1
=
=
= .
2y2
dx
2 1x
2 A - 1x B
and
EXAMPLE 2
Find the slope of the circle x 2 + y 2 = 25 at the point s3, -4d.
Solution The circle is not the graph of a single function of x. Rather it is the combined
graphs of two differentiable functions, y1 = 225 - x 2 and y2 = - 225 - x 2 (Figure
9). The point s3, -4d lies on the graph of y2 , so we can find the slope by calculating the
derivative directly, using the Power Chain Rule:
y
y1 25 x 2
dy2
-6
3
- 2x
`
= `
= = .
4
dx x = 3
2225 - x 2 x = 3
2225 - 9
–5
0
x
5
We can solve this problem more easily by differentiating the given equation of the
circle implicitly with respect to x:
(3, – 4)
y2 –25 x 2
d
- (25 - x 2) 1>2 =
dx
1
- (25 - x 2) -1>2(-2x)
2
Slope – xy 3
4
d 2
d 2
d
(x ) +
(y ) =
(25)
dx
dx
dx
FIGURE 9 The circle combines the
graphs of two functions. The graph of is
y
the lower semicircle and passes through2
s3, -4d .
2x + 2y
dy
= 0
dx
dy
x
= -y.
dx
x
The slope at s3, -4d is - y `
= s3, -4d
3
3
= .
-4
4
Derivatives of Higher Order
Implicit differentiation can also be used to find higher derivatives.
Find d 2y>dx 2 if 2x 3 - 3y 2 = 8 .
To start, we differentiate both sides of the equation with respect to x in order to
EXAMPLE 3
Solution
find y¿ = dy>dx .
d
d
(2x 3 - 3y 2) =
s8d
dx
dx
6x 2 - 6yy¿ = 0
Treat y as a function of x.
2
x
y¿ = y ,
when y Z 0
Solve for y¿.
We now apply the Quotient Rule to find y– .
y– =
Finally, we substitute
2xy - x 2y¿
x2
d
x2
2x
ay b =
= y - 2 # y¿
2
dx
y
y
y¿ = x 2>y to express y– in terms of x and y.
y– =
x2 x2
2x
x4
2x
y - y2 a y b = y - y3 ,
when y Z 0
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Lenses, Tangents, and Normal Lines
Tangent
Light ray
Curve of lens
surface
Normal line
A
Point of entry
P
In the law that describes how light changes direction as it enters a lens, the important angles
are the angles the light makes with the line perpendicular to the surface of the lens at the
point of entry (angles A and B in Figure 3.32). This line is called the normal to the surface
at the point of entry. In a profile view of a lens like the one in Figure 10, the normal is the
line perpendicular to the tangent of the profile curve at the point of entry.
B
Show that the point (2, 4) lies on the curve x 3 + y 3 - 9xy = 0. Then
find the tangent and normal to the curve there (Figure 11).
EXAMPLE 5
FIGURE 10 The profile of a lens,
showing the bending (refraction) of a ray
of light as it passes through the lens
surface.
Solution The point (2, 4) lies on the curve because its coordinates satisfy the equation
given for the curve: 23 + 43 - 9s2ds4d = 8 + 64 - 72 = 0.
To find the slope of the curve at (2, 4), we first use implicit differentiation to find a
formula for dy>dx:
x 3 + y 3 - 9xy = 0
d 3
d 3
d
d
(x ) +
(y ) (9xy) =
(0)
dx
dx
dx
dx
y
t
en
ng
a
T
3x 2 + 3y 2
4
y3
al
rm
No
x3
9xy 0
dy
dy
dx
- 9 ax
+ y b = 0
dx
dx
dx
s3y 2 - 9xd
dy
+ 3x 2 - 9y = 0
dx
3s y 2 - 3xd
0
2
x
FIGURE 11 Example 5 shows how to
find equations for the tangent and normal
to the folium of Descartes at (2, 4).
Differentiate both sides
with respect to x.
Treat xy as a product and y
as a function of x.
dy
= 9y - 3x 2
dx
3y - x 2
dy
= 2
.
dx
y - 3x
Solve for dy>dx.
We then evaluate the derivative at sx, yd = s2, 4d:
dy
3y - x 2
3s4d - 22
8
4
= .
`
= 2
`
= 2
=
5
10
dx s2, 4d
y - 3x s2, 4d
4 - 3s2d
The tangent at (2, 4) is the line through (2, 4) with slope 4>5:
4
sx - 2d
5
4
12
.
y = x +
5
5
y = 4 +
The normal to the curve at (2, 4) is the line perpendicular to the tangent there, the line
through (2, 4) with slope -5>4:
y = 4 y = -
5
sx - 2d
4
5
13
.
x +
4
2