MA 3972-MA-Book May 9, 2018 10:9
126 STEP 4. Review the Knowledge You Need to Score High
[–2, 4] by [–2, 4]
Figure 7.4-2
TIP • Remember that the lim
x→ 0
sin 6x
sin 2x
=
6
2
=3 because the limx→ 0
sinx
x
=1.
7.5 Derivatives of Inverse Functions
Let f be a one-to-one differentiable function with inverse function f−^1 .If
f′(f−^1 (a))=0, then the inverse function f−^1 is differentiable ata and (f−^1 )′(a)=
1
f′(f−^1 (a))
. (See Figure 7.5-1.)
f′ (f–^1 (a))
1
(a, f–^1 (a))
y
0 x
y = x
f
f–^1
(f–^1 (a), a)
m = (f–^1 )′(a)
m = f′(f–^1 (a))
(f–^1 )′^ (a) =
Figure 7.5-1
Ify=f−^1 (x) so thatx=f(y), then
dy
dx
=
1
dx/dy
with
dx
dy
=0.
Example 1
Iff(x)=x^3 + 2 x−10, find (f−^1 )′(x).
Step 1: Check if (f−^1 )′(x) exists. f′(x)= 3 x^2 +2 and f′(x)>0 for all real values ofx.
Thus, f(x) is strictly increasing which implies that f(x)is1−1. Therefore,
(f−^1 )′(x) exists.
Step 2: Lety=f(x) and thusy=x^3 + 2 x−10.
Step 3: Interchangexandyto obtain the inverse functionx=y^3 + 2 y−10.
Step 4: Differentiate with respect toy:
dx
dy
= 3 y^2 + 2.
Step 5: Apply formula
dy
dx
=
1
dx/dy
.
dy
dx
=
1
dx/dy
=
1
3 y^2 + 2
.Thus, (f−^1 )′(x)=
1
3 y^2 + 2