MA 3972-MA-Book May 9, 2018 10:9
Differentiation 127Example 2
Example 1 could have been done by using implicit differentiation.
Step 1: Lety=f(x), and thusy=x^3 + 2 x−10.
Step 2: Interchangexandyto obtain the inverse functionx=y^3 + 2 y−10.
Step 3: Differentiate each term implicitly with respect tox.
d
dx
(x)=
d
dx
(y^3 )+
d
dx
(2y)−
d
dx(−10)
1 = 3 y^2
dy
dx+ 2
dy
dx− 0
Step 4: Solve for
dy
dx
.
1 =
dy
dx
(3y^2 +2)dy
dx=
1
3 y^2 + 2
.Thus, (f−^1 )′(x)=1
3 y^2 + 2.
Example 3
Iff(x)= 2 x^5 +x^3 +1, find (a) f(1) and f′(1) and (b) (f−^1 )(4) and (f−^1 )′(4).
Entery 1 = 2 x^5 +x^3 +1. Sincey1 is strictly increasing, f(x) has an inverse.
(a) f(1) =2(1)^5 +(1)^3 + 1 = 4
f′(x)= 10 x^4 + 3 x^2
f′(1)=10(1)^4 +3(1)^2 = 13(b) Sincef(1)=4 implies the point (1, 4) is on the curvef(x)= 2 x^5 +x^3 +1, therefore,
the point (4, 1) (which is the reflection of (1, 4) ony=x) is on the curve (f−^1 )(x).
Thus, (f−^1 )(4)=1.
(f−^1 )′(4)=1
f′(1)=
1
13
Example 4
Iff(x)= 5 x^3 +x+8, find (f−^1 )′(8).
Entery 1 = 5 x^3 +x+8. Sincey1 is strictly increasing nearx=8,f(x) has an inverse near
x=8.
Note thatf(0)=5(0)^3 + 0 + 8 =8, which implies the point (0, 8) is on the curve of f(x).
Thus, the point (8, 0) is on the curve of (f−^1 )(x).
f′(x)= 15 x^2 + 1
f′(0)= 1Therefore, (f−^1 )′(8)=
1
f′(0)