Math 311 W08 Day 16
Section 4.1 The Algebra of Derivatives
1. Suppose you had a function that was continuous on the interval (a, b)
and you wanted to figure out the slope of the graph at the point x0 in
(a, b). How would you figure it out? Remember that you don’t know
how to take derivatives yet – and probably I would give you one that
was VERY hard to differentiate anyways.
2. Terminology: An interval (a, b) that contains the number x0 is called a
neighborhood of x0.
3. Definition: Let I be a neighborhood of x0. Then the function f: I → R
f (x) " f (x0 )
is said to be differentiable at x0 provided that lim
exists
x!x0
x " x0
in which case we denote this limit by f ′(x0).
If the function is differentiable at every point in I, we say f is
differentiable and call the function f ′: I → R the derivative of f.
4. Determine the values for which each of the following functions are
differentiable. If the functions are differentiable on their entire
domain, then determine the derivative f ′:
a. f: R → R Given by f(x) = x2 + 5x for all x.
b. f: R → R Given by f(x) = x for all x.
Proposition 4.5 (Differentiable functions are continuous). Let I be a
neighborhood of x0 and suppose that the function f: I → R is
differentiable at x0. Then f is continuous at x0.
Proof. Since f is differentiable at x0, we know that
f (x) " f (x0 )
lim
exists. Clearly lim x " x0 exists (it is zero), so by the
x!x0
x!x0
x " x0
product property of limits, lim f (x) " f (x0 ) exists and is zero. Then by
x!x0
the sum property, lim f (x) = f (x0 ) √
x!x0
5. Theorem 4.6 Let I be a neighborhood of x0 and suppose that the
functions f: I → R and g: I → R, are differentiable at x0. Then
i. The sum f + g: I → R is differentiable at x0 and
(f + g)′(x0,) = f ′(x0) + g′(x0,) Proof:
ii. The product fg: I → R is differentiable at x0 and
(fg)′(x0,) = f (x0) g′(x0,) + f ′(x0)g(x0,) Proof:
iii. If g(x) ≠ 0 for all x in I, then the reciprocal 1/g: I → R is
! 1$'
( g '(x0 )
differentiable at x0 and # & (x0 ) =
Proof:
" g%
(g(x0 ))2
iv. If g(x) ≠ 0 for all x in I, then the quotient f/g: I → R is differentiable
! f $'
g(x0 ) f '(x0 ) ( f (x0 )g '(x0 )
at x0 and # & (x0 ) =
Proof:
" g%
(g(x0 ))2
4.2 Differentiating Inverses and Compositions
1. Suppose you have a function that is invertible and differentiable at a
point x0. Is its inverse differentiable at the point f(x0)? If so, what is
the derivative of the inverse?
a. First try to come up with a conjecture by drawing a graph of
an invertible differentiable function. Pick a point, draw a
tangent line to it, and then think about the picture of the
inverse function.
b. Test your conjecture by playing with a couple of examples.
2. Theorem 4.11 Let I be a neighborhood of x0 and let the function
f: I → R be strictly monotone and continuous. Suppose that f is
differentiable at the point x0 and that f′(x0) ≠ 0. Define J = f(I). Then
the inverse f-1: J → R is differentiable at the point y0 = f (x0) and
1
( f !1 )"(y0 ) =
f "(x0 )
6. Proof: First note that J is a neighborhood of y0 by IVT. We know that
since f is continuous, on I, f-1 is continuous on J (Theorem 3. 29), so
we know that lim f "1 (y) = f "1 (y0 ) .
y!y0
f (x) " f (x0 )
= f ′(x0). Then by the
x!x0
x " x0
composition property for limits (Theorem 3.37), (plugging the inverse
function into the difference quotient) we have that
We also know that lim
f ( f "1 (y)) " f ( f "1 (y0 ))
lim
= f ′(x0). But,
y!y0
f "1 (y) " f "1 (y0 )
f ( f "1 (y)) " f ( f "1 (y0 ))
y " y0
lim
lim
=
. So we have
y!y0
y!y0 f "1 (y) " f "1 (y )
f "1 (y) " f "1 (y0 )
0
y " y0
lim "1
= f ′(x0).
y!y0 f
(y) " f "1 (y0 )
y ! y0
will be
f !1 (y) ! f !1 (y0 )
nonzero, and we are given that its limit f ′(x0) is nonzero. Thus we may use
the quotient property of limits to conclude that
1
f "1 (y) " f "1 (y0 )
!1
lim
= ( f )"(y0 ) =
Coda.
y!y0
f "(x0 )
y " y0
Finally observe that as long as y ≠ y0, the function
7. Verify this theorem by comparing the derivatives of the following
functions with derivatives of their inverses:
a. f(x) = 2x
b. f(x) = x2
c. f(x) = ex