It happens that if f is a differentiable one to one function defined on an interval,
[a,b]
,
and f^{′}
(x)
exists and is non zero then the inverse function f^{−1} has a derivative at the
point f
(x )
. Recall that f^{−1} is defined according to the formula
f−1(f (x)) = x.
Let f :
[a,b]
→ ℝ be a continuous function. Recall from Theorem 5.2.1
f (x )− f (a) f (x )− f (b)
f′(a) ≡ lim -----------,f′(b) ≡ lim -----------.
x→a+ x − a x→b− x − b
Recall the notation x → a+ means that only x > a are considered in the definition of
limit, the notation x → b− defined similarly. Thus, this definition includes the derivative
of f at the endpoints of the interval and to save notation,
f (x)− f (x1)
f′(x1) ≡ xli→mx1---x−-x-----
1
where it is understood that x is always in
[a,b]
.
Theorem 5.3.1Let f :
[a,b]
→ ℝ be continuousand one to one. Supposef^{′}
(x1)
exists for some x_{1}∈
[a,b]
and f^{′}
(x1)
≠0. Then
(f −1)
^{′}
(f (x1))
exists andis given by the formula,
(f−1)
^{′}
(f (x ))
1
=
--1--
f′(x1)
.
Proof:By Lemma 4.4.3, and Corollary 4.5.1 on Page 282f is either strictly increasing
or strictly decreasing and f^{−1} is continuous on an interval f
|| x − x 1 ||
||-------1---− -′---|| < ε.
f (x)− f (x1) f (x1)
It follows that if 0 <
|f (x1)− f (x)|
< η,
| | | |
||f−1(f-(x-))−-f−-1(f (x1)) --1---|| ||---x−-x1---- --1--||
| f (x)− f (x1) − f′(x1)| = |f (x)− f (x1) − f′(x1)| < ε
Therefore, since ε > 0 is arbitrary, for y ∈ f
([a,b])
,
− 1 −1
lim f---(y)−-f--(f (x1))= -′1--■
y→f(x1) y − f (x1) f(x1)
The following obvious corollary comes from the above by not bothering with end
points.
Corollary 5.3.2Let f :
(a,b)
→ ℝ, where −∞ ≤ a < b ≤ ∞ be continuousand one to one. Suppose f^{′}
(x1)
exists for some x_{1}∈
(a,b)
and f^{′}
(x1)
≠0. Then
( −1)
f
^{′}
(f (x1))
exists and is given by the formula,
( −1)
f
^{′}
(f (x1))
=
--1--
f′(x1)
.
This is one of those theorems which is very easy to remember if you neglect
the difficult questions and simply focus on formal manipulations. Consider the
following.
−1
f (f (x)) = x.
Now use the chain rule on both sides to write
( −1)′ ′
f (f (x)) f (x) = 1,
and then divide both sides by f^{′}
(x)
to obtain
(f −1)′(f (x)) =-1--.
f′(x)
Of course this gives the conclusion of the above theorem rather effortlessly and it is
formal manipulations like this which aid in remembering formulas such as the one given
in the theorem.
Example 5.3.3Let f
(x)
= 1 + x^{2} + x^{3} + 7. Show that f has an inverse and find
( )
f −1
^{′}
(8)
.
I am not able to find a formula for the inverse function. This is typical in useful
applications so you need to get used to this idea. The methods of algebra are insufficient
to solve hard problems in analysis. You need something more. The question is to
determine whether f has an inverse. To do this,
f′(x) = 2x+ 3x2 + 7 > 0
By Corollary 5.9.5 on Page 379, this function is strictly increasing on ℝ and so it has an
inverse function although I have no idea how to find an explicit formula for this inverse
function. However, I can see that f
(0)
= 8 and so by the formula for the derivative of an
inverse function,
( −1)′ ( −1)′ 1 1
f (8) = f (f (0)) = f′(0) = 7 .
In practice, we typically don’t bother with the mathematical details. We have
f
(x )
= y and the inverse function is of the form x = f^{−1}
(y)
. Thus it involves finding
dx
dy
(y)
≡
(f−1)
^{′}
(y)
. The existence of the derivative of the inverse function exists by the
above argument. Therefore, all that remains is to use the chain rule. Take