** The Derivative**

**4.4 Inverse function theorem**

Note: less than 1 lecture (optional section, needed for , requires )

### 4.4.1

### Inverse function theorem

Let us start with a simple example. Consider a function f(x):=ax for a numbera_{6}=0. Then
f: R→Ris bijective, and the inverse is f−1(y) =1_{a}y. In particular, f0(x) =aand(f−1)0(y) =1_{a}. As

differentiable functions are “infinitesimally like” linear functions, then we expect the same behavior
from the inverse function. The main idea of differentiating inverse functions is the following lemma.
Lemma 4.4.1. Let I,J _{⊂}R be intervals. If f: I →J is strictly monotone (hence one-to-one),

onto ( f(I) =J), differentiable at x0∈I, and f0(x0)6=0, then the inverse f−1 is differentiable at y0= f(x0)and

(f−1)0(y0) = _{f}_{0} _{f}_{−}1_{1}_{(y}

0) =

1
f0_{(x}_{0}_{)}.

If f is continuously differentiable and f0_{is never zero, then f}−1_{is continuously differentiable.}
Proof. By , f has a continuous inverse. For convenience call the inverseg: J→I.
Letx0,y0be as in the statement. For anyx∈I writey:= f(x). Ifx6=x0and soy6=y0, we find

g(y)_{−}g(y0)
y_{−}y0 =
g f(x)
−g f(x0)
f(x)_{−} f(x0) =
x_{−}x0
f(x)_{−}f(x0).
See for the geometric idea.

x=g(y) x0=g(y0) f(x) =y f(x0) =y0 slope= f(x)−f(x0) x−x0 = y−y0 g(y)−g(y0) f(x) =y f(x0) =y0 x=g(y) x0=g(y0) slope= x−x0 f(x)−f(x0)= g(y)−g(y0) y−y0

Figure 4.10:Interpretation of the derivative of the inverse function.

Let
Q(x):=
( x_{−}x_{0}
f(x)_{−}f(x0) ifx6=x0,
1
f0_{(}_{x}_{0}_{)} ifx=x0 (notice that f0(x0)6=0).

As f is differentiable atx0, we have
lim
x→x0Q(x) =xlim→x0
x−x0
f(x)_{−}f(x0) =
1
f0_{(x}_{0}_{)} =Q(x0),

that is,Qis continuous atx0. Asg(y)is continuous aty0, the compositionQ g(y)

= g(y)−g(y0)
y−y0 is
continuous aty0by . Therefore
1
f0 _{g(y}_{0}_{)} =Q g(y0)
= lim
y→y0Q g(y)
= lim
y→y0
g(y)−g(y0)
y_{−}y0 .
Sogis differentiable aty0andg0(y0) = _{f}0_{(}_{g}1_{(}_{y}_{0}_{)}_{)}.

If f0 _{is continuous and nonzero at all}_{x}_{∈}_{I}_{, then the lemma applies at all}_{x}_{∈}_{I}_{. As}_{g}_{is also}
continuous, the derivativeg0_{(y) =} 1

f0_{(}_{g}_{(}_{y}_{)}_{)} must be continuous.

What is usually called the inverse function theorem is the following result.

Theorem 4.4.2 (Inverse function theorem). Let f: (a,b)_{→}R be a continuously differentiable

function, x0∈(a,b)a point where f0(x0)6=0. Then there exists an open interval I⊂(a,b)with x0∈I, the restriction f|I is injective with a continuously differentiable inverse g: J→I defined on an interval J:= f(I), and

g0_{(y) =} 1

f0 _{g(y)}, for all y∈J.

Proof. Without loss of generality, suppose f0_{(x}_{0}_{)}_{>}_{0. As} _{f}0_{is continuous, there must exist an open}
intervalI= (x0−δ,x_{0}+δ)such that f0(x)>0 for allx∈I. See .

By f is strictly increasing onI, and hence the restriction f_{|I} bijective onto
J:= f(I). As f is continuous, then by the (or directly via the

) f(I)is in interval. Now apply .

If you tried to prove the existence of roots directly as in you saw how difficult that endeavor is. However, with the machinery we have built for inverse functions it becomes an almost trivial exercise, and with the lemma above we prove far more than mere existence.

Corollary 4.4.3. Given any n_{∈}_{N}and any x _{≥}0 there exists a unique number y_{≥}0 (denoted
x1/n_{:}_{=}_{y), such that y}n_{=}_{x. Furthermore, the function g}_{:} _{(}_{0}_{,}_{∞}_{)}_{→}_{(}_{0}_{,}_{∞}_{)}_{defined by g(x)}_{:}_{=}_{x}1/n

is continuously differentiable and

g0_{(x) =} 1
nx(n−1)/n =

1

nx(1−n)/n,
using the convention xm/n_{:}_{= (x}1/n_{)}m_{.}

Proof. Forx=0 the existence of a unique root is trivial.

Let f: (0,∞)_{→}(0,∞)be defined by f(y):=yn. The function f is continuously differentiable
and f0_{(y) =}_{ny}n−1_{, see} _{. For} _{y}_{>}_{0 the derivative} _{f}0 _{is strictly positive and so}
again by , f is strictly increasing (this can also be proved directly). Given any

M>1, f(M) =Mn≥M, and given any 1>ε >0, f(ε) =εn≤ε. For any xwithε <x<M

we have by the that x_{∈} f [ε,M]⊂ f (0,∞). AsM and ε were

arbitrary, f is onto (0,∞), and hence f is bijective. Letgbe the inverse of f and we obtain the
existence and uniqueness of positiventh roots. saysghas a continuous derivative and
g0_{(x) =} 1

f0_{(}_{g}_{(}_{x}_{)}_{)}=

1
n(x1/n_{)}n−1.

Example 4.4.4: The corollary provides a good example of where the inverse function theorem
gives us an interval smaller than(a,b). Take f: _{R}_{→}_{R}defined by f(x):=x2. Then f0(x)_{6}=0 as
long asx_{6}=0. Ifx0>0, we can takeI= (0,∞), but no larger.

Example 4.4.5: Another useful example is f(x):=x3. The function f: R→Ris one-to-one and

onto, so f−1_{(y) =}_{y}1/3_{exists on the entire real line including zero and negative} _{y}_{. The function}
f has a continuous derivative, but f−1_{has no derivative at the origin. The point is that} _{f}0_{(}_{0}_{) =}_{0.}
See for a graph, notice the vertical tangent on the cube root at the origin. See also

.

y=x1/3 y=x3

Figure 4.11:Graphs ofx3andx1/3.

### 4.4.2

### Exercises

Exercise4.4.1: Suppose f:R→Ris continuously differentiable such that f0(x)>0for all x. Show that f

is invertible on the interval J= f(R), the inverse is continuously differentiable, and(f−1)0(y)>0for all

y∈ f(R).

Exercise4.4.2: Suppose I,J are intervals and a monotone onto f:I_{→}J has an inverse g:J_{→}I. Suppose

you already know that both f and g are differentiable everywhere and f0_{is never zero. Using chain rule but}

not prove the formula g0_{(}_{y}_{) =} 1

f0_{(}g(y)).

Exercise4.4.3: Let n∈Nbe even. Prove that every x>0has a unique negative nth root. That is, there

exists a negative number y such that yn_{=}_{x. Compute the derivative of the function g}_{(}_{x}_{)}_{:}_{=}_{y.}

Exercise4.4.4: Let n_{∈}Nbe odd and n≥3. Prove that every x has a unique nth root. That is, there exists a

number y such that yn_{=}_{x. Prove that the function defined by g}_{(}_{x}_{)}_{:}_{=}_{y is differentiable except at x}_{=}_{0}_{and}

Exercise4.4.5(requires ): Show that if in the inverse function theorem f has k continuous derivatives,

then the inverse function g also has k continuous derivatives.

Exercise4.4.6: Let f(x):=x+2x2_{sin}_{(}_{1}_{/}_{x}_{)}_{for x}_{6}_{=}_{0}_{and f}_{(}_{0}_{) =}_{0. Show that f is differentiable at all x,}
that f0_{(}_{0}_{)}_{>}_{0, but that f is not invertible on any open interval containing the origin.}

Exercise4.4.7:

a) Let f:R→Rbe a continuously differentiable function and k>0be a number such that f0(x)≥k for

all x∈R. Show f is one-to-one and onto, and has a continuously differentiable inverse f−1:R→R.

b) Find an example f:R→Rwhere f0(x)>0for all x, but f is not onto.

Exercise4.4.8: Suppose I,J are intervals and a monotone onto f:I→J has an inverse g:J→I. Suppose
x∈I and y:= f(x)_{∈}J, and that g is differentiable at y. Prove:

a) If g0_{(}_{y}_{)}_{6}_{=}_{0, then f is differentiable at x.}