Advanced Calculus and Analysis MA1002 - Ian Craw

g.
5.4. Proposition. Let f and g be differentiable at a, and let k be a constant. Then k:f ,
f + g and fg are differentiable at f . Also, if g(a) 6= 0, then f=g is differentiable at a. Let
f be differentiable at a, and let g be differentiable at f(a). Then g  f is differentiable at a.
In addition, the usual rules for calculating these derivatives apply.
5.5. Example. Let f(x) = tan

x2 − a2
x2 + a2

for a 6= 0. Show that f is dierentiable at every
point of its domain, and calculate the derivative at each such point.
Solution. This is the same example we considered in 4.20. There we showed the domain
was the whole of R, and that the function was continuous everywhere. Let g(x) =
x2 − a2
x2 + a2
.
Then g is properly dened for all values of x, and the quotient is dierentiable, since each
term is, and since x2 + a2 6= 0 for any x since a 6= 0. Thus f is dierentiable using the
chain rule since f = tan g, and we are assuming known that the elementary functions like
tan are dierentiable.
Finally to actually calculate the derivative, we have:-
f 0(x) = sec2

x2 − a2
x2 + a2

:
(x2 + a2):2x − ((x2 − a2):2x)
(x2 + a2)2
=
4a2x
(x2 + a2)2
: sec2

x2 − a2
x2 + a2

:
5.6. Exercise. Let f(x) = exp

1 + x2
1− x2

. Show that f is dierentiable at every point of its
domain, and calculate the derivative at each such point.
The rst point in our study of dierentiable functions is that it is more restrictive for
a function to be dierentiable, than simply to be continuous.
5.7. Proposition. Let f be differentiable at a. Then f is continuous at a.
5.2. SIMPLE LIMITS 43
Proof. To establish continuity, we must prove that limx!a f(x) = f(a). Since the Newton
quotient is known to converge, we have for x 6= a,
f(x)− f(a) = f(x)− f(a)
x− a :(x− a) ! f
0(a):0 as x ! a.
Hence f is continuous at a.
5.8. Example. Let f(x) = jxj; then f is continuous everywhere, but not dierentiable at 0.
Solution. We already know from Example 4.15 that jxj is continuous. We compute the
Newton quotient directly; recall that jxj = x if x  0, while jxj = −x if x < 0. Thus
lim
x!0+
f(x)− f(0)
x− 0 = limx!0+
x− 0
x− 0 = 1; while limx!0−
f(x)− f(0)
x− 0 = limx!0−
−x− 0
x− 0 = −1:
Thus both of the one-sided limits exist, but are unequal, so the limit of the Newton quotient
does not exist.
5.2 Simple Limits
Our calculus of dierentiable functions can be used to compute limits which otherwise prove
troublesome.
5.9. Proposition (l’Hoˆpital’s rule: simple form). Let f and g be functions such that
f(a) = g(a) = 0 while f 0(a) and g0(a) both exits and g0(a) 6= 0. Then
lim
x!a
f(x)
g(x)
=
f 0(a)
g0(a)
:
Proof. Since f(a) = g(a) = 0, provided x 6= a, we have
f(x)
g(x)
=
f(x)− f(a)
g(x)− g(a) =
f(x)− f(a)
x− a
x− a
g(x) − g(a) !
f 0(a)
g0(a)
as x ! a;
where the last limit exists, since g0(a) 6= 0.
5.10. Remark. If f 0(a) and g0(a) exist, computing lim
x!a
f(x)
g(x)
is easy by 4.16, since f and g
must be continuous at a by Proposition 5.7, unless we get an indeterminate form 0=0 or
1=1 for the formal quotient. In fact l’Ho^pitals rule helps in both cases, although we need
to develop stronger forms.
5.11. Example. Show that lim
x!0
3x− sinx
x
= 2.
Solution. Note rst that we cannot get the result trivially from 4.16, since since g(a) = 0
and so we get the indeterminate form 0=0. However, we are in a position to apply the
simple form of l’Ho^pital, since x0 = 1 6= 0. Applying the rule gives
lim
x!0
3x− sinx
x
= lim
x!0
3− cos x
1
= 2:
44 CHAPTER 5. DIFFERENTIABILITY
5.12. Example. Show that lim
x!0
p
1 + x− 1
x
= 1=2.
Solution. Note again that we cannot get the result trivially from 4.16, since this gives
the indeterminate 0=0 form, because g(a) = 0. However, we are in a position to apply the
simple form of l’Ho^pital, since x0 = 1 6= 0. Applying the rule gives
lim
x!0
p
1 + x− 1
x
= lim
x!0
2−1(1 + x)−1/2
1
= 1=2:
5.13. Exercise. Evaluate lim
x!0
log(1 + x)
sin x
.
5.14. Example. (Spurious, but helps to remember!) Show that lim
x!0
sinx
x
= 1.
Solution. This is spurious because we need the limit to calculate the derivative in the rst
place, but applying l’Ho^pital certainly gives the result.
5.3 Rolle and the Mean Value Theorem
We can combine our denition of derivative with the Intermediate Value Theorem to give
a useful result which is in fact the basis of most elementary applications of the dieren-
tial calculus. Like the results on continuous functions, it is a global result, and so needs
continuity and dierentiability on a whole interval.
5.15. Theorem (Rolle’s Theorem). Let f be continuous on [a; b], and differentiable on
(a; b), and suppose that f(a) = f(b). Then there is some c with a < c < b such that
f 0(c) = 0.
Note: The theorem guarantees that the point c exists somewhere. It gives no indication
of how to nd c. Here is the diagram to make the point geometrically:
x
c
f’(c) = 0
a b
Figure 5.1: If f crosses the axis twice, somewhere between the two crossings, the function
is flat. The accurate statement of this \obvious" observation is Rolle’s Theorem.
Proof. Since f is continuous on the compact interval [a; b], it has both a global maximum
and a global minimum. Assume rst that the global maximum occurs at an interior point
c 2 (a; b). In what follows, we pick h small enough so that c + h always lies in (a; b). Then
5.3. ROLLE AND THE MEAN VALUE THEOREM 45
If h > 0,
f(c + h)− f(c)
h
 0, and so lim
h!0+
f(c + h)− f(c)
h
 0, since we know the limit
exists.
Similarly, if h < 0,
f(c + h)− f(c)
h
 0, and so lim
h!0+
f(c + h)− f(c)
h
 0, since we
know the limit exists. Combining these, we see that f 0(c) = 0, and we have the result in
this case.
A similar argument applies if, instead, the global minimum occurs at the interior point
c. The remaining situation occurs if both the global maximum and global minimum occur
at end points; since f(a) = f(b), it follows that f is constant, and any c 2 (a; b) will do.
5.16. Example. Investigate the number of roots of each of the polynomials
p(x) = x3 + 3x + 1 and q(x) = x3 − 3x + 1:
Solution. Since p0(x) = 3(x2 + 1) > 0 for all x 2 R, we see that p has at most one root;
for if it had two (or more) roots there would be a root of p0(x) = 0 between them by Rolle.
Since p(0) = 1, while p(−1) = −3, there is at least one root by the Intermediate Value
Theorem. Hence p has exactly one root.
We have q0(x) = 3(x2 − 1) = 0 when x = 1. Since q(−1) = 3 and q(1) = −1, there is
a root of q between −1 and 1 by the Intermediate Value Theorem. Looking as x !1 and
as x ! −1 shows here are three roots of q.
5.17. Exercise. Show that the equation x − e−x = 0 has exactly one root in the inter-
val (0; 1).
Our version of Rolle’s theorem is valuable as far as it goes, but the requirement that
f(a) = f(b) is suciently strong that it can be quite hard to apply sometimes. Fortunately
the geometrical description of the result | that somewhere the tangent is parallel to the
axis, does have a more general restatement.
5.18. Theorem (The Mean Value Theorem). Let f be continuous on [a; b], and dif-
ferentiable on (a; b). Then there is some c with a < c < b such that
f(b)− f(a)
b− a = f
0(c) or equivalently f(b) = f(a) + (b− a)f 0(c):
Proof. We apply Rolle to a suitable function; let
h(x) = f(b)− f(x)− f(b)− f(a)
b− a (b− x):
Then h is continuous on the interval [a; b], since f is, and in the same way, it is dierentiable
on the open interval (a; b). Also, f(b) = 0 and f(a) = 0. We can thus apply Rolle’s theorem
to h to deduce there is some point c with a < c < b such that h0(c) = 0. Thus we have
0 = h0(c) = −f 0(c) + f(b)− f(a)
b− a ;
which is the required result.
46 CHAPTER 5. DIFFERENTIABILITY
bc
A
B
a
Figure 5.2: Somewhere inside a chord, the tangent to f will be parallel to the chord. The
accurate statement of this common-sense observation is the Mean Value Theorem.
5.19. Example. The function f satises f 0(x) =
1
5− x2 and f(0) = 2. Use the Mean Value
theorem to estimate f(1).
Solution. We rst estimate the given derivative for values of x satisfying 0 < x < 1. Since
for such x, we have 0 < x2 < 1, and so 4 < 5− x2 < 5. Inverting we see that
1
5
< f 0(x) <
1
4
when 0 < x < 1:
Now apply the Mean Value theorem to f on the interval [0; 1] to obtain some c with 0 < c < 1
such that f(1)− f(0) = f 0(c). From the given value of f(0), we see that 2:2 < f(1) < 2:25
5.20. Exercise. The function f satises f 0(x) =
1
5 + sin x
and f(0) = 0. Use the Mean
Value theorem to estimate f(=2).
Note the \common sense" description of what we have done. If the derivative doesn’t
change much, the function will behave linearly. Note also that this gives meaning to the
approximation
f(a + h)  f(a) + hf 0(a):
We now see that the accurate version of this replaces f 0(a) by f 0(c) for some c between a
and a + h.
5.21. Theorem. (The Cauchy Mean Value Theorem) Let f and g be both continuous
on [a; b] and differentiable on (a; b). Then there is some point c with a < c < b such that
g0(c)

f(b)− f(a)

= f 0(c)

g(b) − g(a)

:
5.4. L’HOˆPITAL REVISITED 47
In particular, whenever the quotients make sense, we have
f(b)− f(a)
g(b)− g(a) =
f 0(c)
g0(c)
:
Proof. Let h(x) = f(x)

g(b) − g(a)

− g(x)

f(b) − f(a)

, and apply Rolle’s theorem
exactly as we did for the Mean Value Theorem. Note rst that since both f and g are
continuous on [a; b], and dierentiable on (a; b), it follows that h has these properties. Also
h(a) = f(a)g(b)− g(a)f(b) = h(b). Thus we may apply Rolle to h, to deduce there is some
point c with a < c < b such that h0(c) = 0. But
h0(c) = f 0(c)

g(b) − g(a)

− g0(c)

f(b)− f(a)

Thus
f 0(c)

g(b) − g(a)

= g0(c)

f(b)− f(a)

This is one form o