James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

154 Chapter 2 Limits and Derivatives
2
f
_2 2
_2
2
fª
_2 2
ExamplE 2
(a) If f sxd − x 3 2 x, find a formula for f 9sxd.
(b) Illustrate this formula by comparing the graphs of f and f 9.
SOLUtion
(a) When using Equation 2 to compute a derivative, we must remember that the variable
is h and that x is temporarily regarded as a constant during the calculation of the limit.
f 9sxd − lim
h l 0
f sx 1 hd 2 f sxd
h
− lim
h l 0
fsx 1 hd 3 2 sx 1 hdg 2 fx 3 2 xg
h
− lim
h l 0
x 3 1 3x 2 h 1 3xh 2 1 h 3 2 x 2 h 2 x 3 1 x
h
− lim
h l 0
3x 2 h 1 3xh 2 1 h 3 2 h
h
− lim
h l 0
s3x 2 1 3xh 1 h 2 2 1d − 3x 2 2 1
FIGURE 3
_2
(b) We use a graphing device to graph f and f 9 in Figure 3. Notice that f 9sxd − 0
when f has horizontal tangents and f 9sxd is positive when the tangents have positive
slope. So these graphs serve as a check on our work in part (a).
n
ExamplE 3 If f sxd − sx , find the derivative of f. State the domain of f 9.
SOLUtion
f 9sxd − lim
h l0
f sx 1 hd 2 f sxd
h
y
− lim
h l0
sx 1 h 2 sx
h
− lim
h l0S sx 1 h 2 sx
h
? sx 1 h 1 sx
sx 1 h 1 sx
D
(Rationalize the numerator.)
1
0
1
x
− lim
h l0
sx 1 hd 2 x
h(sx 1 h 1 sx ) − lim
h l0
h
h(sx 1 h 1 sx )
y
(a) ƒ=œ„x
− lim
h l0
1
sx 1 h 1 sx
−
1
− 1
sx 1 sx 2sx
1
We see that f 9sxd exists if x . 0, so the domain of f 9 is s0, `d. This is slightly smaller
than the domain of f , which is f0, `d.
n
0
1
1
(b) fª(x)=
2œ„x
FIGURE 4
x
Let’s check to see that the result of Example 3 is reasonable by looking at the graphs
of f and f 9 in Figure 4. When x is close to 0, sx is also close to 0, so f 9sxd − 1y(2sx )
is very large and this corresponds to the steep tangent lines near s0, 0d in Figure 4(a) and
the large values of f 9sxd just to the right of 0 in Figure 4(b). When x is large, f 9sxd is very
small and this corresponds to the flatter tangent lines at the far right of the graph of f and
the horizontal asymptote of the graph of f 9.
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