Rules of Differentiation

Rules of Differentiation

The process of finding the derivative of a function is called Differentiation.1In the previous chapter, the required derivative of a function is worked out by taking the limit of the difference quotient. It would be tedious, however, to have to do this every time we wanted to find the derivative of a function, for there are various rules of differentiation that will enable us to find the derivative of a desired function directly. Students are advised to equip themselves with the following rules to be able to apply them in the subsequent topics like marginal analysis, optimisation (Unconstrained and Constrained) problems.

Following are some of the rules of Differentiation.

1. Constant Function Rule The derivative of a constant function 𝑓(𝑥) = 𝐴 , where A is a constant, is zero.

I.e. if 𝑓(𝑥) = 𝐴 then 𝑓′(𝑥) = 0

The reason for 𝑓′(𝑥) = 0 for 𝑓(𝑥) = 𝐴 is easy to see intuitively by having a look at the graph (Fig 2.1) of a constant function. The graph is a horizontal straight line with a zero Slope throughout.

Example 2.1. Given 𝑓(𝑥) = −5

𝑓′(𝑥) = 0

𝑓(𝑥) = 10

𝑓′(𝑥) = 0

1 Differentiation is an operation that transforms a function 𝑓 into another function𝑓′.

2. Linear Function Rule The derivative of a linear function 𝑓(𝑥) = 𝑎 + 𝑏𝑥 with 𝑎 and 𝑏 constants is equal to 𝑏. For example the derivative of a function 𝑓(𝑥) = 3 + 2𝑥 is 2.it is obvious that the derivative of a linear function is the multiplicative constant of the variable. The important implication of this result is that for a linear function the rate at which variable y changes with respect to a change in 𝑥 is same at every value of 𝑥.

Example 2.2. Find the derivative of

a. 𝑞 = 10 − 2𝑝 ; where q is quantity demanded and p the price b. 𝐶 = 5 + 1 𝑌 ; where C is consumption , Y is income
2

Solution a. Given 𝑞 = 3 − 2𝑝 ⇒

𝑑𝑞 = −2 means that quantity demanded falls by two
𝑑𝑝

units for every one unit increase in the price ( see Fig. 2.2) .

a. Given 𝐶 = 5 + 1 𝑌 ⇒ 𝑑𝑐 = 1

2

𝑑𝑦 2

The derivative 𝑑𝑐 or 𝐶′ is called marginal propensity to consume (mpc). In the above example
𝑑𝑦
the value of mpc = 0.5 > 0 . the graphical exposition of the function is shown in fig.2.3.

3. Power Function Rule. The derivative of a function 𝑓(𝑥) = 𝑥𝑛; where n is any arbitrary constant is 𝑛 multiplied by the variable raised to the power 𝑛 − 1. i.e if 𝑓(𝑥) = 𝑥𝑛 then
𝑓′(𝑥) = 𝑛𝑥𝑛−1.
For 𝑛 = 2, the rule was already confirmed in the example 10 of the previous chapter. Few remarks about the power function 𝑓(𝑥) = 𝑥𝑛
 If 𝑛 > 1; the derivative of the function will be increasingfunction of 𝑥  If 𝑛 = 1; the drivative of the function will remain constant function and will be equal to 1  If 𝑛 < 1; the derivative of the function will be decreasing function of 𝑥 .

Example 2.3. find the derivative of the following functions given below and draw there respective Graphs to illustrate the remarks about power function.

3
a. 𝑓(𝑥) = 𝑥2

b. 𝑓(𝑥) = 𝑥

1
c. 𝑓(𝑥) = 𝑥2

3
Solution. a). Given 𝑓(𝑥) = 𝑥2 b) Given 𝑓(𝑥) = 𝑥

⇒

𝑓′(𝑥) = 3 𝑥32−1 =

3

1
𝑥2

2

2

⇒ 𝑓′(𝑥) = 1

[∵ 𝑥 = 𝑥1 ]

1
c) Given 𝑓(𝑥) = 𝑥2

⇒ 𝑓′(𝑥) = 1 𝑥12−1 = 1 𝑥−12 = 1

2

2

2√𝑥

the diagramatic exposition of the above functions is shown in Fig. 2.4.2

2 If x is considered as Labour input, then the three graphs portrays increasing, constant and decreasing marginal physical productivity of Labour.

3.1. Generalised Power Function Rule
The generalised power function is symbolically written as 𝑓(𝑥) = 𝑔(𝑥)𝑛 where 𝑔 is a differentiable function and 𝑛 is any real number. The derivative of such function is
𝒇′(𝒙) = 𝒏[𝒈(𝒙)]𝒏−𝟏. 𝒈′(𝒙)
The above result shows, while differentiating power function, multiply the original index (𝑛) by 𝑔(𝑥) raised to the power diminished by unity and also by the derivative 𝑔′(𝑥).

Example 2.4. Differentiate by applying power function rule to following functions

a. 𝑓(𝑥) = 𝑥−12 b. 𝑓(𝑥) = 3(𝑥)4 c. 𝑓(𝑥) = (𝑥2 + 3𝑥4)5 d. 𝑓(𝑥) = √𝑥 Solution a) Given 𝑓(𝑥) = 𝑥−12 ⇒ 𝑓′(𝑥) = −12𝑥−12−1 = −12𝑥−13
b) Given 𝑓(𝑥) = 3𝑥4 ⇒ 𝑓′(𝑥) = 3 .4. 𝑥4−1 = 12𝑥3. c) Given 𝑓(𝑥) = (𝑥2 + 3𝑥4)5 Here 𝑔(𝑥) = (𝑥2 + 3𝑥4) ⇒ 𝑔′(𝑥) = 2𝑥 + 12𝑥3

∴ 𝑓′(𝑥) = 5 (𝑥2 + 3𝑥4)5−1. (2𝑥 + 12𝑥3)

= 5(𝑥2 + 3𝑥4)4. ( 2𝑥 + 12𝑥3)

1
d) Given 𝑓(𝑥) = √𝑥 = 𝑥2

⇒

𝑓′(𝑥) = 1 𝑥12−1 =

1

𝑥

−1
2

2

2

4. The derivative of Sum and Difference.

The derivative of sum (difference) of two functions is equal to the sum (difference) of their derivatives. i.e.

if 𝑓(𝑥) = 𝑔(𝑥) + ℎ(𝑥), then 𝑓′(𝑥) = 𝑔′(𝑥) + ℎ′(𝑥) and (∗)

if 𝑓(𝑥) = 𝑔(𝑥) − ℎ(𝑥), then 𝑓′(𝑥) = 𝑔′(𝑥) − ℎ′(𝑥)

(∗∗)

The rule set out above for the case of a combination of two functions only. But, if more than two functions appear in a combination, the rules can be applied several times in succession to give the derivative. It can be noticed, however, that the sum and difference rule extends at once to give the derivative of the algebraic sum of a number of functions as the similar algebraic sum of the derivatives of the separate functions. For example,
If 𝑓(𝑥) = 𝑔(𝑥) + ℎ(𝑥) + 𝑤(𝑥) , then 𝑓′(𝑥) = 𝑔′(𝑥) + ℎ′(𝑥) + 𝑤(𝑥)

And if 𝑓(𝑥) = 𝑔(𝑥) − ℎ(𝑥) − 𝑤(𝑥) , then 𝑓′(𝑥) = 𝑔′(𝑥) − ℎ′(𝑥) − 𝑤(𝑥) The important deductions from (∗) and (∗∗) concerns the behaviour of constants in the process of derivation. A constant (additive) can be regarded as the case of a function of 𝑥 which does not change in value as 𝑥 varies. This means that the derivative of constant is zero. Hence an additive constant disappears when the derivative is taken.3 To better under as to why constant if added to a function disappears on differentiating.4 Consider two functions 𝑓(𝑥) = 𝑥2 and 𝑓(𝑥) = 𝑥2 + 𝐴, where A is a constant and plot them the figure that emerges (fig 2.5) clearly reveals that the slope (derivative) of the functions at every point is same, the only difference that is apparent is that the graph of the later function is A steps away from the graph of the former function.
Example 2.5. Find the derivative of the following functions. a. 𝑓(𝑥) = 3𝑥4 + 2𝑥3 − 𝑥2 + 10 b. 𝑓(𝑥) = 3√𝑥 +2𝑥 − 14 c. 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 ; where 𝑎, 𝑏, 𝑐 are constants
Solution. a) Given 𝑓(𝑥) = 3𝑥4 + 2𝑥3 − 𝑥2 + 10
3 Multiplicative constants remain unaffected by taking the derivative of a function. If 𝑓(𝑥) = 2𝑥4 ; then 𝑓′(𝑥) = 2 𝑓′(𝑥4) = 8𝑥3
4 This fact provides mathematical explanation of the well known economic principle that fixed costs of a firm does not affect its marginal cost.

𝑓′(𝑥) = 3𝑓′(𝑥4) + 2𝑓′(𝑥3) − 𝑓′(𝑥2) = 3 .4 𝑥3 + 2 .3 𝑥2 − 2𝑥 = 12 𝑥3 + 6𝑥2 − 2𝑥

b) Given 𝑓(𝑥) = 3√𝑥 +2𝑥 − 14

3
𝑓′(𝑥) = 𝑓′(𝑥)2 + 2𝑓′(𝑥)

=

3

(

𝑥

)

1 2

+ 2

2

c) Given 𝑓(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

𝑓′(𝑥) = 𝑎𝑓′(𝑥2) + 𝑏𝑓′(𝑥)

= 2𝑎𝑥 + 𝑏

5 product function Rule

If 𝑓(𝑥) = ℎ(𝑥). 𝑔(𝑥), then

𝑓′(𝑥) = ℎ(𝑥). 𝑔′(𝑥) + 𝑔(𝑥). ℎ′(𝑥)

In words: the derivative of the product of two functions is equal to the derivative of the first function times the derivative of the second function plus the second function times the derivative of first function. The significant point to remember about the product function rule is that the derivative of the product of two functions is not the simple product of their separate derivatives i.e. 𝑓′(𝑥) ≠ 𝑔′(𝑥). ℎ′(𝑥). For example if ℎ(𝑥) = 𝑥2 and 𝑔(𝑥) = 4𝑥 then (ℎ. 𝑔). (𝑥) = ( 𝑥2). (4𝑥) = 4𝑥3 it is obvious here that (ℎ. 𝑔)′(𝑥) = 12𝑥2 is not equal to ℎ′(𝑥). 𝑔′(𝑥) = (2𝑥). (4) = 8𝑥.
Verification by Product Rule: if 𝑓(𝑥) = (𝑥)2. (4𝑥)
Consider ℎ(𝑥) = (𝑥)2 and 𝑔(𝑥) = 4𝑥 then
By applying product rule
𝑓′(𝑥) = ℎ(𝑥). 𝑔′(𝑥) + 𝑔(𝑥). ℎ′(𝑥)
= 𝑥2(4) + 4𝑥(2𝑥) = 4𝑥2 + 8𝑥2 = 12𝑥2
The same result was obtained when we first multiply the two functions and then differentiate the result directly. The obvious question that emerges is that why do we need product rule when we

have another option of multiplying the functions and taking the derivative of the product directly. One response to the query is that multiplying first and then differentiating is applicable when the functions are given in some specific form, but the product rule is applicable even when the functions are given in general form.5

Example 2.6. Find the derivative of the following functions

a. 𝑓(𝑥) = 3𝑥4(2𝑥5 + 5𝑥) b. 𝑓(𝑥) = (√𝑥)(3√𝑥 + 2) solution. a) Given 𝑓(𝑥) = 3𝑥4(2𝑥5 + 5𝑥)

c. 𝑓(𝑥) = (𝑥4 + 1) ( 𝑥3 + 2)
𝑥

Let 𝑔(𝑥) = 3𝑥4 and ℎ(𝑥) = (2𝑥5 + 5𝑥)

Then 𝑔′(𝑥) = 12𝑥3 and ℎ′(𝑥) = (10𝑥4 + 5).plug in these values in product rule formula

𝑓′(𝑥) = 𝑔(𝑥). ℎ′(𝑥) + ℎ(𝑥). 𝑔′(𝑥) = 3𝑥4(10𝑥4 + 5). (2𝑥5 + 5𝑥). (12𝑥3) = 54𝑥8 + 75𝑥4

b) Given 𝑓(𝑥) = (√𝑥)(3√𝑥 + 2)

Let 𝑔(𝑥) = (√𝑥) and ℎ(𝑥) = 3√𝑥 + 2)

Differentiating 𝑔(𝑥) and ℎ(𝑥) separately and substitute the values in the product rule formula

𝑔′(𝑥) = 1

3
and ℎ(𝑥) = (𝑥 + 2)2

⇒

ℎ′(𝑥)

=

3

1
(𝑥)2

2√𝑥

2

∴ 𝑓′(𝑥) = 𝑔(𝑥). ℎ′(𝑥) + ℎ(𝑥). 𝑔′(𝑥)

=

(√𝑥).

3

1
(𝑥)2

+

(𝑥

+

3
2)2

(

1

)=

3𝑥 +

1

3
(𝑥 + 2)2

2

2√𝑥

2

2√𝑥

c) Given 𝑓(𝑥) = (𝑥4 + 1) ( 𝑥3 + 2)
𝑥

Let 𝑔(𝑥) = (𝑥4 + 1) and ℎ(𝑥) = ( 𝑥3 + 2)
𝑥

5 For further exposition see Chiang A. C. Fundamental Methods of Mathematical Economics (4th ed.), McGraw-Hill (2005).

Differentiating 𝑔(𝑥) and ℎ(𝑥) separately and substitute the values in the product rule formula 𝑔′(𝑥) = 4𝑥3 − 𝑥12 and ℎ′(𝑥) = 3𝑥2 ∴ 𝑓′(𝑥) = 𝑔(𝑥). ℎ′(𝑥) + ℎ(𝑥). 𝑔′(𝑥)
= (𝑥4 + 𝑥1) (3𝑥2) + ( 𝑥3 + 2)(4𝑥3 − 𝑥12) 6. Derivative of the Quotient of two Functions (Quotient Rule)
If (𝑥) = 𝑔(𝑥) ; where 𝑔(𝑥) and ℎ(𝑥) are both differentiable at 𝑥 and ℎ(𝑥) ≠ 0 then
ℎ(𝑥)
𝑓′(𝑥) = ℎ(𝑥).𝑔′(𝑥) − 𝑔(𝑥).ℎ′(𝑥)
[ℎ(𝑥)]2

In words: The derivative of a quotient is equal to the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
Note: in the product rule formula, the two functions appear symmetrically, so that it is easy to remember. In the quotient rule however, the expression in the numerator must be in the right order. The simplest way to check whether the order of the expression in the numerator is appropriate or not, imagine that ℎ(𝑥) = 1 so thatℎ′(𝑥) = 0. If on substituting the formula reduces to 𝑔′(𝑥) then the signs are correct and if it reduces to −𝑔′(𝑥), then the signs are wrong.

Example 2.7. Find the derivative of the following functions

a. 𝑓(𝑥) = 𝑥+1 where 𝑥 ≠ 1
𝑥−1

b. 𝑓(𝑥) = 5𝑥2𝑥6+2−3𝑥54

c. 𝑓(𝑥) = 𝑐(𝑥) d. 𝑓(𝑥) = √𝑥+1

𝑥

𝑥

Solution a) Given 𝑓(𝑥) = 𝑥+1 where 𝑥 ≠ 1
𝑥−1

let 𝑔(𝑥) = 𝑥 + 1 and ℎ(𝑥) = 𝑥 − 1 differentiate them individually we have.

𝑔′(𝑥) = 1 and ℎ′(𝑥) = 1 substitute the value in the formula

𝑓′(𝑥) = ℎ(𝑥).𝑔′(𝑥) − 𝑔(𝑥).ℎ′(𝑥)
[ℎ(𝑥)]2

= (𝑥−1)((1𝑥)−−1()𝑥2+1)(1) = 𝑥−(𝑥1−−1𝑥)−2 1 = (𝑥−−21)2

b) Given 𝑓(𝑥) = 52𝑥𝑥6+2−35𝑥4
Let 𝑔(𝑥) = 5𝑥6 + 3𝑥4 and ℎ(𝑥) = 2𝑥2 − 5 differentiate individually we have

𝑔′(𝑥) = 30𝑥5 + 12𝑥3 and ℎ′(𝑥) = 4𝑥 , substitute the values in the formula

𝑓′(𝑥) = ℎ(𝑥).𝑔′(𝑥) − 𝑔(𝑥).ℎ′(𝑥)
[ℎ(𝑥)]2

= (2𝑥2−5).(30𝑥5+12𝑥3)−(5𝑥6+3𝑥4).(4𝑥) = 40𝑥7−138𝑥5−60𝑥3

(2𝑥2−5) 2

(2𝑥2−5) 2

c) Given 𝑓(𝑥) = 𝑐(𝑥)
𝑥

If in the above example 𝑥 is considered as the level of output and 𝑐(𝑥) as the cost of producing 𝑥 units then 𝑓(𝑥) = 𝑐(𝑥) turns out to be general average cost function.
𝑥

Let 𝑔(𝑥) = 𝑐(𝑥) and ℎ(𝑥) = 𝑥 differentiating individually we have

𝑔′(𝑥) = 𝑐′(𝑥) and ℎ′(𝑥) = 1 substitute the values in the formula

𝑓′(𝑥) = ℎ(𝑥).𝑔′(𝑥) − 𝑔(𝑥).ℎ′(𝑥)
[ℎ(𝑥)]2

= 𝑥.𝑐′(𝑥𝑥)−2 𝑐(𝑥) = 𝑥1 [𝑐′(𝑥) − 𝑐(𝑥𝑥) ]..........(∗)

The economic meaning of the expression (∗) is that for positive output levels(𝑥 > 0), the

marginal cost 𝑐′(𝑥)

𝑐(𝑥)

exceeds the average cost

if and only if the rate of change of the

𝑥

average cost as output increase is positive.

a) Given 𝑓(𝑥) = √𝑥+1
𝑥
Let 𝑔(𝑥) = √𝑥 + 1 and ℎ(𝑥) = 𝑥 differentiating 𝑔 and ℎ individually we have 𝑔′(𝑥) = 1 √𝑥 and ℎ′(𝑥) = 1 substitute the values in the formula
2
𝑓′(𝑥) = ℎ(𝑥).𝑔′(𝑥) − 𝑔(𝑥).ℎ′(𝑥)
[ℎ(𝑥)]2