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Section 5.
- Use the definition to find the derivative of each of the following functions.
f. g(x) =
x x^2 +
, x ∈ R
g
′ (x) = lim h→ 0
g(x + h) − g(x)
h
= lim h→ 0
x+h (x+h)^2 +
x x^2 +
h
= lim h→ 0
−h(x^2 −1+xh) ((x+h)^2 +1)(x^2 +1)
h
= lim h→ 0
x
2 − 1 + xh
((x + h)^2 + 1) (x^2 + 1)
1 − x
2
(x
2
2
- For each of the following, determine whether the given function is differentiable at the indicated
point x 0. Justify your answer!
a. f (x) = x|x|, at x 0 = 0.
f is differentiable at x = 0. Since for x ≥ 0, f (x) = x
2 and for x < 0, f (x) = −x
2 ,
f
′ (0) = lim h→ 0 −
f (0 + h) − f (0)
h
= lim h→ 0 −
−h
2
h
= lim h→ 0 −^
−h = 0
and
f
′ +(0)^ =^ lim h→ 0 +
f (0 + h) − f (0)
h
= lim h→ 0 +
h
2
h
= lim h→ 0 +^
h = 0.
Therefore f
′ (0) = f
′ (0) = f
′ +(0) = 0.^2
b. f (x) =
x
2 x ∈ Q
0 , x /∈ Q
at x 0 = 0
f is differentiable at x = 0. For h 6 = 0, define function
g(h) =
f (0 + h) − f (0)
h
f (h)
h
We show that limh→ 0 g(h) = 0. For any ε > 0, let δ = ε. When 0 < |h| < δ, if h ∈ Q, then
|g(h) − 0 | =
f (h)
h
h
2
h
= |h| < δ = ε,
if h /∈ Q,
|g(h) − 0 | =
f (h)
h
h
= 0 < ε.
Therefore f
′ (0) = limh→ 0 g(h) = 0. 2
c. g(x) = (x − 2)[x], at x 0 = 2,
For 1 < x < 2, g(x) = (x − 2). For 2 < x < 3, g(x) = 2(x − 2). So g
′ −(2) = 1 and^ g
′ +(2) = 2. Therefore
g is not differentiable at x = 2. 2
d. h(x) =
x + 2.
h is not differentiable at x = −2. Note that Dom h = [− 2 , ∞). So there is no left derivative h
′ −(−2).
In fact there is not right derivative h
′ +(−2) either.
h
′ +(−2)^ =^ lim x→− 2 +
h(x) − h(−2)
x − (−2)
= lim x→− 2 +
x + 2
x + 2
= lim x→− 2 +
x + 2
e. f (x) =
sin x x ∈ Q
x, x /∈ Q
at x 0 = 0
f is differentiable at x = 0. For h 6 = 0, define function
g(h) =
f (0 + h) − f (0)
h
f (h)
h
We show that limh→ 0 g(h) = 1. For any ε > 0, since
lim h→ 0
sin h
h
There exists a δ 1 > 0 such that when 0 < |h| < δ 1 ,
sin h
h
− 1 | < ε.
Let δ = min{ε, δ 1 }. When 0 < |h| < δ, if h ∈ Q, then 0 < |h| < δ 1 ,
|g(h) − 1 | =
f (h)
h
sin h
h
< ε,
if h /∈ Q,
|g(h) − 1 | =
f (h)
h
h
h
∣ = 0^ < ε.
Therefore f
′ (0) = limh→ 0 g(h) = 1. 2
- Let g be defined by g(x) =
x
2 sin
1 x x^6 = 0 0 , x = 0
a. Prove that g is differentiable at 0 and that g
′ (0) = 0.
Proof. Sine
lim x→ 0
g(x) − g(0)
x − 0
= lim x→ 0
x
2 sin
1 x
x
= lim x→ 0
x sin
x
g is differentiable at 0 and that g
′ (0) = 0. 2
b. Show that g
′ (x) is not continuous at 0.
Proof. Using the Product Rule and Chain Rule, we have
g
′ (x) = 2 x sin
x
2
cos
x
x
2
= 2 x sin
x
− cos
x
