Class notes Differential Calculus

Module 4: DERIVATIVE OF A FUNCTION

4.1 Increment and Derivatives

Let y be a continued-single function of x denoted by y = f (x) . If the independent variable x changes from
x to x + x , then the dependent variable y will also change from y to y + y ; thus y = f (x) changes to
y + y = f ( x + x) . Hence 𝛥𝑦 = 𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥).

If the variable x changes from one fixed value x1 to another value x 2 the difference x2 − x1 is called the
increment of x, denoted by x .
It should be noted that the increment may be positive or negative.

Example If y = x2 – 3x + 1, compute y when
(a) x changes from x1 = 1 to x2 = 1.1; (b) x changes from x1 = –1 to x2 = –1.1

Solution:
Solving for y, y = (x + x)2 – 3(x + x) + 1 – (x2 – 3x + 1) = 2xx + (x)2 – 3x = x(2x +x – 3)
(a) when x changes from x1 = 1 to x2 = 1.1 then x = 0.1, so y = (0.1)(2 + 0.1– 3) = –0.09
(b) when x changes from x1 = –1 to x2 = –1.1 then x = –0.1, so y = (–0.1)( –2–0.1– 3) = 0.51

f ( x + x)
Consider the function y = f(x). If for a fixed value of x, the quotient approaches a limit as the
x
increment x approaches zero, the limit is called the derivative of y with respect to x for the given value of x.
dy
This is usually denoted by .
dx

The derivative of y with respect to x is defined as
dy y f ( x + x) − f ( x)
= lim = lim
dx x→0 x x→0 x
provided the limit exists

Other notations for derivatives are:
d
Dx y, Dx f ( x), y' , f ' ( x), and f ( x)
dx

The process of finding the derivative is called differentiation. If the derivative exists, f(x) is said to be a
differentiable function.

One method of determining the derivative of a function is the increment method or more commonly known as
the four-step rule. The procedure is as follows.

Step 1 Substitute x + x for x and y + y for y in y = f(x).
Step 2 Subtract y = f(x) from the result of step 1 to obtain y in terms of x and x .
Step 3 Divide both sides of step 2 by x .
Step 4 Find the limit of step 3 as x approaches zero.

𝑑𝑦
Example 1 If y = x2 – 3x + 1, find
𝑑𝑥

Solution:
Step 1 → y + y = (x + x)2 – 3(x + x) + 1
Step 2 → y = (x + x)2 – 3(x + x) + 1 – (x2 – 3x + 1) = 2xx + (x)2 – 3x = x(2x +x – 3)
𝑦 𝑥(2𝑥 +𝑥 – 3)
Step 3 → 𝑥 = = 2𝑥 + 𝑥 – 3
𝑥
𝑦
Step 4 → lim = lim (2𝑥 + 𝑥 – 3) = 2𝑥 − 3
𝑥→0 𝑥 𝑥→0
𝑑𝑦

, 𝑑𝑦 3𝑥−2
Example 2 Determine given that 𝑦 = 2𝑥+1
𝑑𝑥

Solution:
3(𝑥+∆𝑥)−2
Step 1 → y + y = 2(𝑥+∆𝑥)+1
3(𝑥+∆𝑥)−2 3𝑥−2 (2𝑥+1)(3𝑥+3∆𝑥−2)−(3𝑥−2)(2𝑥+2∆𝑥+1) 7∆𝑥
Step 2 → y = 2(𝑥+∆𝑥)+1 − =
(2𝑥+2∆𝑥+1)(2𝑥+1)
= (2𝑥+2∆𝑥+1)(2𝑥+1)
2𝑥+1
𝑦 7∆𝑥 7
Step 3 → 𝑥 = ∆𝑥(2𝑥+2∆𝑥+1)(2𝑥+1) = (2𝑥+2∆𝑥+1)(2𝑥+1)
𝑦 7 7
Step 4 → lim 𝑥 = lim ((2𝑥+2∆𝑥+1)(2𝑥+1)) = (2𝑥+1)2
𝑥→0 𝑥→0
𝑑𝑦 7
 =
𝑑𝑥 (2𝑥+1)2


dy
Example 3. Evaluate for y = x − 1 when x =17.
dx
Solution:
Step 1 → y + y = √𝑥 + ∆𝑥 − 1
Step 2 → y = √𝑥 + ∆𝑥 − 1 − √𝑥 − 1
𝑦 √𝑥+∆𝑥−1−√𝑥−1 √𝑥+∆𝑥−1−√𝑥−1 √𝑥+∆𝑥−1+√𝑥−1 (𝑥+∆𝑥−1)−(𝑥−1) 1
Step 3 → = = ( ) = ∆𝑥(√𝑥+∆𝑥−1+√𝑥−1) =
𝑥 ∆𝑥 ∆𝑥 √𝑥+∆𝑥−1+√𝑥−1 √𝑥+∆𝑥−1+√𝑥−1
𝑦 1 1
Step 4 → lim = lim ( ) = 2√𝑥−1
𝑥→0 𝑥 𝑥→0 √𝑥+∆𝑥−1+√𝑥−1

𝑑𝑦 1
When x = 17, 𝑑𝑥 = 8


4.2 Rules of Algebraic Differentiation


Rule 1. Derivative of a Constant

d
If c is any constant, and y is a function defined by y = c then c=0
dx
dy
Example Determine if 𝑦 = 32 + 𝑎𝑟𝑐𝑠𝑖𝑛5
dx

Solution:
dy
Since  is a mathematical constant and arcsin5 is also a constant, then =0
dx
Rule 2. Derivative of a Power
If n is any real number and if y = u n , where u is a differentiable function of x, then

u n = nu n −1 (u )
d d
dx dx
dy
Example Find if 𝑦 = 𝑥 4
dx

Solution:
dy
Applying rule 2 or also known as power formula gives, = 4𝑥 3
dx

Rule 3. Derivative of a Constant times a Function
If u is a differentiable function of x and C is a constant, then
d
(Cu) = C d (u ) .
dx dx