18.014 Problem Set 6 Solutions
Sam Elder
October 21, 2015
Problem 1 (20 points). The following functions are defined on R. For each function, determine at which
points in R the function is differentiable and compute the derivative at those points.
√
Part 1.1. f (x) = 1 + x2 .
Answer. This function is differentiable everywhere, with f 0 (x) = √
x
.
1 + x2
√
Solution. Let g(x) = x and h(x) = 1 + x2 , so we can write f = g ◦ h. We know that h is differentiable
1
everywhere with h0 (x) = 2x and g is differentiable on x > 0 with g 0 (x) = 2√
. Since h(x) = 1 + x2 ≥ 1 > 0
x
for all x, we can apply the chain rule to get
2x
x
f 0 (x) = g 0 (h(x))h0 (x) = √
=√
.
2
2 1+x
1 + x2
Part 1.2. f (x) = x · |x|.
Answer. This function is differentiable everywhere, with f 0 (x) = 2|x|.
Solution. Let g(x) = x and h(x) = |x|. We know that both of these are continuous on x 6= 0, with g 0 (x) = 1
and h0 (x) = |x|
x . Therefore, by the product rule, when x 6= 0,
f 0 (x) = g(x)h0 (x) + h(x)g 0 (x) = x ·
|x|
+ |x| · 1 = 2|x|,
x
as desired. It remains to show that f 0 (0) = 0, which we do by hand. For any h 6= 0, we have
h|h| − 0|0|
f (h) − f (0)
=
= |h|.
h
h
The function |x| is continuous at 0, so limh→0
f 0 (0) = 0 = |0| as desired.
f (h)−f (0)
h
= limh→0 |h| = 0. Since the limit exists, we have
Part 1.3. f (x) = xm (x + 1)n , where m, n ∈ N.
Answer. This function is differentiable everywhere, with f 0 (x) = ((m + n)x + m)xm−1 (x + 1)n−1 .
Solution. Let g(x) = xm , h(x) = xn , and s(x) = x + 1, so f (x) = g(x) · h(s(x)). We have already established
that g(x) = mxm−1 , h(x) = nxn−1 , and s(x) = 1. By the product rule and chain rule, therefore,
f 0 (x) = g(x)h0 (s(x))s0 (x) + g 0 (x)h(s(x)) = xm (n(x + 1)n−1 ) + mxm−1 (x + 1)n
= (nx + m(x + 1))xm−1 (x + 1)n−1 = ((m + n)x + m)xm−1 (x + 1)n−1 .
√
Part 1.4. f (x) = (1 + 3 x)3 .
1
Answer. This function is not differentiable at x = 0. For x 6= 0, f 0 (x) = (1 + x−1/3 )2 .
√
Solution. First, consider x 6= 0. Letting g(x) = (1 + x)3 and h(x) = 3 x, we have f (x) = g(h(x)). We’ll need
this fact about h:
Lemma. h(x) is differentiable on all x 6= 0 with h0 (x) =
1
.
3x2/3
Proof. We have already established this for x > 0; it remains to be shown for x < 0. For x < 0, we
have h(x) = −h(−x) so if s(x) = −x, h(x) = s(h(s(x))). Now s(x) is differentiable as a polynomial with
s0 (x) = −1, so the chain rule (applied twice) establishes
h0 (x) = s0 (h(s(x)))h0 (s(x))s0 (x) = (−1)
1
1
(−1) = 2/3 .
3(−x)2/3
3x
We also need g 0 (x). Letting t(x) = 1 + x and u(x) = x3 , we have g(x) = u(t(x)) so by the chain rule,
g (x) = u0 (t(x))t0 (x) = 3(1 + x)3 · 1 = 3(1 + x)3 .
Now by the chain rule, for x 6= 0,
√
√
(1 + 3 x)2
3(1 + 3 x)2
=
= (1 + x−1/3 )2 .
f 0 (x) = g 0 (h(x))h0 (x) =
√
2
3
3x2/3
x
0
Finally, for x = 0, we claim that the derivative does not exist. Indeed, we can expand the cube with Binomial
Theorem: f (x) = 1 + 3x1/3 + 3x2/3 + x. Then
f (h) − f (0)
1 + 3h1/3 + 3h2/3 + h − 1
=
= 3h−2/3 + 3h−1/3 + 1.
h
h
(0)
= L. Then for = L, there is some δ such
Suppose for sake
of contradiction
that limh→0 f (h)−f
h
f (h) − f (0)
f
(h)
−
f
(0)
− L < L, so
< 2|L|. Take some 0 < h < min{δ, |L|−3 }, so
that if |h| < δ, h
h
f (h) − f (0)
> 3|L|2 + 3|L| + 1 > 2|L|, a contradiction. So indeed, this limit does not exist, as desired.
h
Problem 2. Let n be a positive integer and let f1 , . . . , fn : S → R be differentiable functions with the same
domain. Let g = f1 · · · fn be the product of these functions. Prove that for any x such that g(x) 6= 0, we
have
g 0 (x)
f 0 (x)
f 0 (x)
= 1
+ ··· + n
.
g(x)
f1 (x)
fn (x)
Solution. We induct on n.
Base Case. For n = 1, we have g = f1 , so g 0 (x) = f10 (x). If g(x) = f1 (x) 6= 0, we can divide by it to get
f 0 (x)
g 0 (x)
= 1
,
g(x)
f1 (x)
as desired.
Inductive Step. Suppose that the statement holds for n = m ≥ 1 and we will show it holds for n = m + 1, so
0
let g = f1 f2 · · · fm+1 . Then let h = f1 f2 · · · fm , so g = hfm+1 . By the product rule, g 0 = hfm+1
+ h0 fm+1 .
If g(x) = h(x)fm+1 (x) 6= 0, we can divide by it to get
0
h(x)fm+1
(x) + h0 (x)fm+1 (x)
f 0 (x) h0 (x)
g 0 (x)
=
= m+1
+
.
g(x)
h(x)fm+1 (x)
fm+1 (x)
h(x)
We also know that if h(x)fm+1 (x) =
6 0, h(x) 6= 0. Now by the induction hypothesis, since h = f1 f2 · · · fm ,
whenever h(x) 6= 0,
h0 (x)
f 0 (x) f20 (x)
f 0 (x)
= 1
+
+ ··· + m
h(x)
f1 (x) f2 (x)
fm (x)
2
Plugging this into the equation above yields
f 0 (x)
g 0 (x)
f 0 (x) f20 (x)
= 1
+
+ · · · + m+1
,
g(x)
f1 (x) f2 (x)
fm+1 (x)
as desired.
Induction is complete.
Problem 3. Let f : R → Z be an integer-valued function. Show that if f is differentiable at x ∈ R, then
f 0 (x) = 0.
Solution 1. We get our hands dirty with limits. Suppose for sake of contradiction
that f is differentiable
at
f (x + h) − f (x)
0
0
0
0
x but f (x) 6= 0. Let = |f (x)|, so there is some δ such that if |h| < δ, − f (x) < |f (x)|.
h
(x)
In particular, this means that f (x+h)−f
6= 0, so f (x + h) 6= f (x). As both f (x), f (x + h) ∈ Z,
h
this implies that |f (x + h) − f (x)| ≥ 1 for all |h| < δ. Now take some 0 < h < min{δ, 2|f 01(x)| }. Then
f (x + h) − f (x) ≥ 2|f 0 (x)|, so by the triangle inequality,
h
f (x + h) − f (x)
0
− f (x) ≥ |f 0 (x)|,
h
a contradiction. So indeed, if f : R → Z is differentiable at x, f 0 (x) = 0.
Solution 2. A cleaner solution using continuity. If f is differentiable at x ∈ R, then f is continuous at x.
Taking = 1, there exists δ such that if |y − x| < δ, |f (y) − f (x)| < 1. But since f (x) and f (y) are integers,
their difference is an integer, hence 0. So f is constant near x, and f 0 (x) = 0.
Problem 4. Suppose f : R → R is an even function (that is, f (−x) = f (x) for all x ∈ R) and is differentiable
at 0. Prove that f 0 (0) = 0.
Solution. Let g(x) = −x, and notice that the even property can be written as f (g(x)) = f (x). Since f
is differentiable everywhere, we can take the derivative of both sides. By the chain rule, we get f 0 (x) =
f 0 (g(x))g 0 (x) and since g 0 (x) = −1, we have f 0 (x) = −f 0 (−x). Plugging in x = 0 gives f 0 (0) = −f 0 (0), so
f 0 (0) = 0.
Problem 5.
Part 5.1. Construct a differentiable function f : R → R such that f (x) = 0 for x ≤ 0, f (x) = 1 for x ≥ 1,
and 0 < f (x) < 1 for 0 < x < 1.
Solution. Let1


0
f (x) = 3x2 − 2x3


1
x≤0
0<x<1
x ≥ 1.
This clearly satisfies f (x) = 0 for x ≤ 0 and f (x) = 1 for x ≥ 1. For 0 < x < 1, we have 0 < x3 < x2 <
x < 1, so f (x) = 3x2 − 2x3 > 3x2 − 2x2 = x2 > 0 and 1 − (3x2 − 2x3 ) = (1 − x)(1 + x − 2x2 ) > 0, or
f (x) = 3x2 − 2x3 < 1, as desired.
1 To construct this function, I knew that its derivative needed to equal 0 at both 0 and 1, so I picked a polynomial whose
derivative was a multiple of x(x − 1) = x2 − x.
3
Finally, it remains to show it is differentiable. On (0, 1), f (x) = g(x) is a polynomial, and therefore
differentiable, with g 0 (x) = 6x − 6x2 . Moreover, at x = 0,
g(h) − g(0)
f (h) − f (0)
= lim
= g 0 (0) = 6 · 0 − 6 · 02 = 0,
+
h
h
h→0
0−0
f (h) − f (0)
= lim
= 0.
lim
h
h
h→0+
h→0−
lim
h→0+
Therefore, since the left and right limits agree, limh→0
at x = 1,
f (h)−f (0)
h
exists and f is differentiable at 0. Finally,
f (1 + h) − f (1)
g(1 + h) − g(1)
= lim−
= g 0 (1) = 6 · 1 − 6 · 12 = 6 − 6 = 0,
h
h
h→0
f (h + 1) − f (1)
1−1
lim+
= lim+
= 0,
h
h
h→0
h→0
lim
h→0−
Since the left and right limits agree, limh→0
f (h+1)−f (1)
h
exists and f is differentiable at 1, as desired.
Part 5.2. Use the function f constructed in the previous part to define a new differentiable function
g : R → R such that g(x) = 0 for |x| ≥ 1, g(0) = 1, and 0 < g(x) < 1 for 0 < |x| < 1.
Solution. Let g(x) = f (x + 1)f (1 − x), where f is the piecewise-defined function from the previous part.2
Since f is differentiable and the functions x 7→ x + 1 and x 7→ 1 − x are also differentiable as polynomials, g is
differentiable by the product rule. Then for |x| ≥ 1, either x ≤ −1 or x ≥ 1, so f (x + 1) = 0 or f (1 − x) = 0.
In either case, g(x) = 0 as desired. We also have g(0) = f (1)2 = 12 = 1 as desired.
Finally, if 0 < x < 1, then x + 1 ≥ 1 so f (x + 1) = 1 and 0 < 1 − x < 1 so 0 < f (1 − x) < 1, and hence
0 < g(x) < 1. If −1 < x < 0, 0 < x + 1 < 1, so 0 < f (x + 1) < 1 and 1 − x ≥ 1 so f (1 − x) = 1, and hence
0 < g(x) < 1. Having shown that 0 < g(x) < 1 whenever 0 < |x| < 1, we are done.
Part 5.3. Let h : R → R be a continuous function that is differentiable at all points except 0. Prove that
there exists a function j : R → R, differentiable on all of R, such that j(x) and h(x) have the same sign
(both positive, both negative, or both zero) for all nonzero x and j(x) = h(x) for |x| ≥ 1.
Solution. Let j(x) = h(x)(1 − g(x)). As the product of two differentiable functions (1 − g is the difference of
two differentiable functions), j is differentiable at all points except 0. Moreover, since g(x) < 1 for all nonzero
x, 1 − g(x) > 0 for all nonzero x, and so j(x) and h(x) have the same sign for all nonzero x. Additionally,
for |x| ≥ 1, g(x) = 0, so j(x) = h(x) as desired.
Finally, we verify that j is differentiable at 0. We know that h(x) is continuous at 0, and 1 − g(x) is
differentiable. Since g(0) = 1, we have
j(x) − j(0)
h(x)(1 − g(x))
g(x) − g(0)
−0=
= h(x)
.
x
x
x
Both h(x) and
g(x)−g(0)
x
have a limit as x → 0, so the limit of their product is the product of the limits:
j(x) − j(0)
g(x) − g(0)
lim
− 0 = lim h(x)
lim
= h(0)g 0 (0),
x→0
x→0
x→0
x
x
so j(x) is differentiable at x = 0, as desired.
2 We choose this function in order to make the product always 0 outside of (−1, 1) and 1 at x = 0. Imagine flipping the
negative part of the x-axis around to cover the intervals (−∞, −1) and translating it to cover (1, ∞).
4