Mathvengers • Fall 2022 • Calculus I
Problem Set #3 • Differentiation
Differentiation Rules
Power rule
1. Differentiate the following functions with respect to x.
a) f (x) = x3 − 4x + 6
b) f (x) = 1.4x5 − 2.5x2 + 6.7
c) f (x) = 2x− 4
d) f (x) =
3
1
2
e) f (x) = − + 3x − 2x 2
x
g) f (x) =
√
4
x − 4xe
f) f (x) = xπ +
1
x
√
10
1 + x + x2 + x3
x3
Product rule
2. Differentiate the following functions with respect to x.
b) f (x) = (x2 − 1)(3x3 + 2)
a) f (x) = (x + 3)(2x + 5)
2
c) f (x) = (4x + 5)
1
−x
x
d) f (x) =
e) f (x) = (x + x)(x − x)
2
g) f (x) =
2
3x2 + x + 1
x
f) f (x) =
1
1
+ 2
x x
1
2x − √
x
(x−3 + 1)
Quotient rule
3. Differentiate the following functions with respect to x.
a) f (x) =
3x + 2
2x2 − 1
c) f (x) =
x2
3
+
5−x
5
b) f (x) =
∗
d) f (x) =
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4x2 + 1
x−1
√
x − √1x
x2 + 1
(2x4 + 25)
x−3 − x3
Mathvengers
Problem Set #3
∗
∗
e) f (x) = (x − 3)
2
g) f (x) =
x−1
x+1
∗
√
(2 x + 1)(x − 1)
f) f (x) =
x−3
x−2 − x2
x4 + x + 1
Chain rule
4. Differentiate the following functions with respect to x.
b) f (x) = (3x2 + 2x + 1)−3
a) f (x) = (x2 + 3x)5
c) f (x) =
∗
∗
4 √
+ 3x
x
− 32
d) f (x) =
√
e) f (x) = (x2 + 2x)3 ( x − 1)
g) f (x) =
∗
f) f (x) =
q
√
3
12 + x
√
x3 − 2x + 3
(2x − 1)2
(x2 − 1)3
Mixed problems
5. Suppose f (x) =
√
a2 − x2 where a > 0. Find f
a
2
.
x1012 − 1
.
x→1 x − 1
6. Evaluate lim
d
f (2x) = −4x2 + 6, find f ′ (2).
dx
∗
7. Suppose
∗
8. Evaluate f ′ (0) if f (2x) + f (x) = 4x for all x.
9. Suppose u and v are functions of x that are differentiable at x = 0 and that
• u(0) = 5
• u′ (0) = −3
• v(0) = −1
Find the values of the following derivatives at x = 0.
a)
c)
d
(uv)
dx
d v
dx u
b)
d u
dx v
d)
d
(7v − 2u)
dx
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• v ′ (0) = 2
Mathvengers
∗
Problem Set #3
10. Let f (x) be a polynomial, such that f (1) = 0. If for all x ∈ R, we have
2f (x) − xf ′ (x) − 1 = 0
Find f (x).
11. Let f be a (differentiable) function such that f (x3 − 7) = 3x2 + 5x − 11. Find f ′ (1).
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12. In this problem we want to show that we cannot in general interchange derivatives and
limits. Suppose −1 ≤ x ≤ 1 throughout this question. We consider
fn (x) = x −
(x2
x
+ 1)n
The notation fn (x) reminds us that the function depends on both x and n.
a) Show that lim fn (x) = x for all x ∈ [−1, 1]. (Hint: evaluate the case x = 0 and x ̸= 0
n→∞
separately. For the latter, use squeeze theorem.)
From this part we have that
d
lim fn (x) = 1
dx n→∞
for every x ∈ [−1, 1]
b) Now, differentiate fn (x) with respect to x. We shall call this function fn′ (x).
c) Now, evaluate the expressions from part (a) and (b) at x = 0. If everything went correctly,
we should get
d
d
lim fn (x) = 1
lim
fn (x) = 0
n→∞ dx
dx x=0 n→∞
x=0
However, in practice, we can most probably interchange them freely in the context of
MATH1012. If you are interested in knowing more about interchanging limits and/or
derivatives, feel free to take MATH2033/2043 or consult your tutor. The magic word is
"Uniform Convergence".
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