Math 51- Analytic Geometry with Calculus 1*

Bachelor of Science in Civil Engineering

Surigao Del Sur State University
WEEKS 1-2
November 22, 2021

“A boat that is not tied up will drift along with the stream.”– Japanese Proverbs

5 Differentiation of Algebraic Functions

5.1 Theorems on Differentiation
Definition 5.1. The process of finding the derivative is called differentiation.

In this section we establish some theorems which help us in the process of differentiation so
that we do not have to use the three-step rule.

Theorem 5.2. If f (x) = c, a constant, for all x, then f ′ (x) = 0 for all x. (The derivative
of a constant is zero).

Theorem 5.3. If n is a positive integer and f (x) = xn , then f ′ (x) = nxn−1 .

Theorem 5.4. If f (x) has a derivative f ′ (x), then the derivative of g(x) = cf (x) is cf ′ (x),
where c is any constant. (The derivative of a constant times a function is the con-
stant times the derivative of a function.)

Theorem 5.5. If f (x) and g(x) have derivatives and F (x) = f (x) + g(x), then F ′ (x) =
f ′ (x) + g ′ (x). (The derivative of the sum is the sum of the derivative).

Example 5.6. Find the derivative of

f (x) = 3x5 + 2x4 − 7x2 + 2x + 5.

* Prepared by: Andrew Felix S. Cunanan IV

1
2

Solution: The quantity x5 has the derivative 5x4 , as we know from Theorem 5.3. Using this
fact and Theorem 5.4, we find the derivative of 3x5 to be 15x4 . Similarly, the derivative of

2x4 is 8x3 , −7x2 is −14x, 2x is 2, 5 is 0.

Theorem 5.5 tells us that the derivative of the sum of these terms is the sum of the derivatives.
We obtain
f ′ (x) = 15x4 + 8x3 − 14x + 2.
□

Theorem 5.7. If u(x) and v(x) are any two functions which have a derivative and if f (x) =
u(x) · v(x), then
f ′ (x) = u(x) · v ′ (x) + v(x) · u′ (x).
(The derivative of the product of two functions is the first times the derivative
of the second plus the second times the derivative of the first).

Example 5.8. Find the derivative of

f (x) = (x2 − 3x + 2)(x3 + 2x2 − 6x).

What is the value of f ′ (3)? and the value of f ′ (2)?

Solution: Theorem 5.7 gives us an idea for a simpler way to solve the derivative. We write

u(x) = x2 − 3x + 2 and v(x) = x3 + 2x2 − 6x.

This means that f (x) = u(x)v(x). Now, u′ (x) = 2x − 3 and v ′ (x) = 3x2 + 4x − 6. Then

f ′ (x) = u(x) · v ′ (x) + v(x) · u′ (x)

= (x2 − 3x + 2)(3x2 + 4x − 6) + (x3 + 2x2 − 6x)(2x − 3)
f ′ (3) = (9 − 9 + 2)(27 + 12 − 6) + (27 + 18 − 18)(6 − 3)
= 66 + 81 = 147.
′
f (2) = (4 − 6 + 2)(12 + 8 − 6) + (8 + 8 − 12)(4 − 2)
= 0 + 4 = 4.

□
Theorem 5.9. If u(x) and v(x) are any two functions which have a derivative and if
u(x)
f (x) =
v(x)
with v(x) ̸= 0, then
v(x) · u′ (x) − u(x) · v ′ (x)
f ′ (x) = .
[v(x)]2
 3

(The derivative of a quotient of two functions is the denominator times the

derivative of the numerator minus the numerator times the derivative of the
denominator, all divided by the square of the denominator.)

Example 5.10. Find the derivative of

2x2 − 3x
f 9x) = .
x2 + 3
Solution: We set u(x) = 2x2 − 3x and v(x) = x2 + 3. Then u′ (x) = 4x − 3 and v ′ (x) = 2x.
Now by Theorem 5.9,
(x2 + 3)(4x − 3) − (2x2 − 3x) · 2x 3x2 + 12x − 9
f ′ (x) = = .
(x2 + 3)2 (x2 + 3)2
The last step was merely an algebraic simplification and, strictly speaking, not a part of the
differentiation process. □
Theorem 5.11. If n is a positive integer and f (x) = x−n , with x ̸= 0, then

f (x) = −nx−n−1 .

5.2 The Chain Rule. Applications

The chain rule is one of the most important and useful tool in differentiation.
Theorem 5.12. (Fundamental Lemma of Differentiation). Suppose that F has a
derivative at a value u so that F ′ (u) exists. We define the function

F (u + h) − F (u)
(
G(h) = − F ′ (u), if h ̸= 0
h
0, if h = 0

Then
(a) G is continuous at h = 0, and

(b) the formula

F (u + h) − F (u) = [F ′ (u) + G(h)]h holds.

Theorem 5.13. (Chain Rule). Suppose that f, g, and u are functions with f (x) = g[u(x)],
and suppose that g and u are differentiable. Then f is differentiable and the following formula
holds:
f ′ (x) = g ′ [u(x)] · u′ (x).

Before illustrating the chain rule by examples, we establish one of the most iportant special
cases.
4

Corollary 5.14. If f (x) = [u(x)]n and n is an integer, then

f ′ (x) = n[u(x)]n−1 · u′ (x).

Example 5.15. Find f ′ (x), given that f (x) = (x2 + 3x − 2)4 .

Solution: From the Corollary, with n = 4 and the expression in x being x2 + 3x − 2, we find

f ′ (x) = 4(x2 + 3x − 2)3 · (2x + 3),

since the derivative of x2 + 3x − 2 is 2x + 3. □

Example 5.16. Find dy/dx, given that

1
y= .
x3 + 3x2 − 6x + 4
Solution: We write y = (x3 + 3x2 − 6x + 4)−1 and apply the Corollary with n = −1:
dy
= −1(x3 + 3x2 − 6x + 4)−2 · (3x2 + 6x − 6)
dx
−3(x2 + 2x − 2)
= 3 .
(x + 3x2 − 6x + 4)2

□
We note that this example could also have been worked by using the formula for the derivative
of a quotient.

dy
Example 5.17. Find dx
, given that
 7
3x − 2
y= .
2x + 1
Solution: This requires a combination of formula. First, from the Corollary, we see that
 6  
dy 3x − 2 3x − 2
=7 · derivative of .
dx 2x + 1 2x + 1
To find the derivative of (3x − 2)/(2x + 1) we apply the quotient formula to get
 6
dy 3x − 2 (2x + 1)(3) − (3x − 2)(2)
=7 ·
dx 2x + 1 (2x + 1)2
49(3x − 2)6
= .
(2x + 1)8
□
 5

Example 5.18. Find f ′ (x), given that

f (x) = (x2 + 2x − 3)16 (2x + 5)13 .

Solution: We first observe that if we let u(x) = (x2 + 2x − 3)16 and v(x) = (2x + 5)13 , we
may use Theorem 5.7, on the product of two functions:

f ′ (x) = u(x) · v ′ (x) + v(x) · u′ (x).

To find u′ (x) we employ the chain rule:

u′ (x) = 16(x2 + 2x − 3)15 (2x + 2);

and similarly,
v ′ (x) = 13(2x + 5)12 (2).
Substitutiing in the formula for f ′ (x), we get

f ′ (x) = (x2 + 2x − 3)16 · 26(2x + 5)12 + (2x + 5)13 · 16(x2 + 2x − 3)15 (2x + 2)
= (x2 + 2x − 3)15 · (2x + 5)12 26(x2 + 2x − 3) + 32(2x + 5)(x + 1) .
 

5.3 The Power Function

The function
f (x) = xn ,
where n is some number, is called a power function.

We follow the same rule for the derivative of x1/q as for the derivative of a power function
when the exponent is an integer. The following theorem establishes this fact.

Theorem 5.19. If f (x) = x1/q and q is an integer, then

1
f ′ (x) = x(1/q)−1 .
q

Definition 5.20. A rational number is one which can be written as one integer over the
other. “The number r is rational” means that r = p/q, where p and q are integers.

Corollary 5.21. If f (x) = xr an r is any rational number, then f ′ (x) = rxr−1 .

Corollary 5.22. If f (x) = [u(x)]r and r is rational, then f ′ (x) = r[u(x)]r−1 u′ (x).
6

√
Example 5.23. Given that f (x) = 2 x3 , find f ′ (x).
5

Solution: f (x) = 2x3/5 and, by the rule for rational exponents,

 
′ 3 (3/5)−1 6 −2/5
f (x) = 2 x = x .
5 5

□
√
Example 5.24. Given that f (t) = 3
t2 + 3t + 1, find f ′ (t).

Solution: We write f (t) = (t3 + 3t + 1)1/3 and use Corollary 5.22:

1 t2 + 1
f ′ (t) = (t3 + 3t + 1)−2/3 (3t2 + 3) = 3 . □
3 (t + 3t + 1)2/3

Example 5.25. Given that f (x) = (x + 1)3 (2x − 1)4/3 , find f ′ (x).

Solution: We use the rule for differentiating a product, setting u(x) = (x + 1)3 and v(x) =
(2x − 1)4/3 , so that f (x) = u(x) · v(x). Then

f ′ (x) = (x + 1)3 · v ′ (x) + (2x − 1)4/3 · u′ (x).

By the chain rule for powers we have

4
u′ (x) = 3(x + 1)2 (1) and v ′ (x) = (2x − 1)1/3 (2).
3
Therefore,
8
f ′ (x) = (x + 1)3 (2x − 1)1/3 + 3(2x − 1)4/3 (x + 1)2
3
8
= (x + 1)2 (2x − 10)1/3 (x + 1) + 3(2x − 1)

3
1
= (x + 1)2 (2x − 1)1/3 (26x − 1).
3
□

Example 5.26. Given the function

s
f :s→ √ ,
s2 − 1

find f ′ (s).

Solution: We write
s
f (s) =
(s2 − 1)1/2
 7

and use Theorem 5.9 for the derivative of a quotient:

(s2 − 1)(1) − s 12 (s2 − 1)−1/2 (2s)

f ′ (s) = .
s2 − 1

Multiplying both numerator and denominator by (s2 − 1)1/2 , we get

′ (s2 − 1)1 − s2 (s2 − 1)0 −1

f (s) = 2 3/2
= 2 .
(s − 1) (s − 1)3/2