CHAPTER TWO: DERIVATIVES

.
2.1 Definition of Derivative

Definition 2.1.The tangent line to the curve 𝑦 = 𝑓(𝑥) at the point 𝑝(𝑎. 𝑓(𝑎))
is the line through 𝑝 with slope.
𝑓(𝑥)−𝑓(𝑎)
(1) 𝑚 = lim
𝑥→𝑎 𝑥−𝑎

Provided that this limit exists.

Example 2.1. Find an equation of the tangent line to the parabola 𝑦 = 𝑥 2 at point
𝑝(1,1).
Solution: Here we have 𝑎 = 1 and 𝑓(𝑥 ) = 𝑥 2 so the slope is
𝑓 (𝑥 ) − 𝑓 (1) 𝑥2 − 1 (𝑥 + 1)(𝑥 − 1)
𝑚 = lim = lim = lim
𝑥→1 𝑥−1 𝑥→1 𝑥 − 1 𝑥→1 𝑥−1
= lim (𝑥 + 1) = 1 + 1 = 2
𝑥→1

Using the point-slope form of the equation of a line we find an equation of the
tangent line at (1,1) is
𝑦 − 1 = 2(𝑥 − 1) or 𝑦 = 2𝑥 − 1
There is another expression for the slope of a tangent line that is sometimes easier
to use. If ℎ = 𝑥 − 𝑎 then 𝑥 = ℎ + 𝑎 and so the slope of the secant line 𝑃𝑄 is
(2) 𝑓 (ℎ + 𝑎) − 𝑓(𝑎)
𝑚𝑃𝑄 =
ℎ
3
Example 2.2. Find an equation of the tangent line to the hyperbola 𝑦 = 𝑥 at the

point (3,1)
3
Solution: Let (𝑥 ) = 𝑥 . Then the slope of the tangent at (3,1) is

𝑓 (ℎ + 𝑎) − 𝑓(𝑎) 𝑓 (3 + ℎ) − 𝑓(3) 3⁄ −1
𝑚 = lim = lim = lim ℎ + 3
ℎ→0 ℎ ℎ→0 ℎ ℎ→0 ℎ
−ℎ −1 1
= lim = lim =−
ℎ→0 ℎ(3 + ℎ) ℎ→0 3 + ℎ 3
Therefore, an equation of the tangent at the point (3,1) is
1
𝑦 − 1 = − (𝑥 − 3
3

1
This simplifies to
𝑥 + 3𝑦 − 6 = 0
The hyperbola and its tangent are shown in figure 3.2

DERIVATIVE:

Definition 2.2: The derivative of a function f at 𝑎, denoted by 𝑓 ′ (𝑎) is

𝑓(𝑥)−𝑓(𝑎)
(3) 𝑓 ′ (𝑎) = lim
(3) 𝑥→𝑎 𝑥−𝑎

if this limit exist

Alternatively, can be defined as

𝑓(ℎ+𝑎)−𝑓(𝑎)
𝑓 ′ (𝑎) = lim , if this limit exists.
ℎ→0 ℎ

If the limit in (3) exists, we say that 𝑓 has a derivative of f at 𝑎 that 𝑓 is

differentiable at 𝑎 or that 𝑓 ′ (𝑎) exists.
1
Example 2.3. Let 𝑓(𝑥 ) = 4 𝑥 2 + 1. Find 𝑓 ′ (−1) 𝑎𝑛𝑑 𝑓 ′ (3)

Solution: Using (3) we obtain

1 1
𝑓 (𝑥 ) − 𝑓 (𝑎 ) 𝑥 2 + 1 − (4 𝑎2 + 1)
′( ) 4
𝑓 𝑎 = lim = lim
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎
1
(𝑥 2 − 𝑎2 ) 1 1 1
4
= lim = lim (𝑥 + 𝑎) = lim (𝑥 + 𝑎) = (𝑎 + 𝑎)
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 4 4 𝑥→𝑎 4
𝑎
=
2
1 3
so, 𝑓 ′ (−1) = − and𝑓 ′ (3) =
2 2

Example 2.4. Find an equation of the tangent line to the parabola

𝑦 = 𝑥 2 − 8𝑥 + 9 at (3, −6).
Solution: The derivative of 𝑓 at a number 𝑎 is 𝑓 ′ (𝑎) = 2𝑎 − 8.
Therefore, the slope of the tangent line at (3, −6) is 𝑓 ′ (3) = −2. Thus, an
equation of the tangent line is
𝑦 − (−6) = (−2)(𝑥 − 3)or 𝑦 = −2𝑥
Example 2.5. Find the equation of the tangent line to the graph of 𝑦 = 𝑥 2 + 1 at
the point (2, 5).

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 Solution: First, we will find the slope of the tangent line (that is the derivative of
𝑓 at 2), 𝑓(𝑥 ) = 𝑥 2 + 1
𝑓 (𝑥 ) − 𝑓 (𝑎 ) (𝑥 2 + 1) − (𝑎2 + 1)
𝑓 ′ (𝑎) = lim = lim = lim (𝑥 + 𝑎) = 2𝑎
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎 𝑥→𝑎

So, 𝑓 ′ (2) = 4.
Thus, the equation of the tangent line is
𝑦 − 𝑓(𝑎) = 𝑓 ′ (𝑎)(𝑥 − 𝑎)
𝑦 − 5 = 4(𝑥 − 2)
Which we can write in slope-intercept form as 𝑦 = 4𝑥 − 3.

Activity 2.2 Find the equation of the tangent line to the graph y = 2x 2 − x 3
at the point (2, 0)

Interpretation of the derivative as the slope of a tangent line to the curve

𝑦 = 𝑓(𝑥). In section 2.1 we defined the tangent line to the curve 𝑦 = 𝑓(𝑥) at the
point 𝑝(𝑎, 𝑓(𝑎)) to be the line that passes through 𝑃 and has slope 𝑚 given by
equation (1). Since by Definition 2.2, this is the same as the derivative 𝑓 ′ (𝑎).

Definition 2.3: The tangent line 𝑦 = 𝑓(𝑥) at (𝑎, 𝑓(𝑎)) is the line
through (𝑎, 𝑓(𝑎)) whose slope is equal to 𝑓 ′ (𝑎), the derivative 𝑓
at 𝑎.

Example 2.6. Find the derivative of the function 𝑓 (𝑥 ) = 𝑥 2 − 8𝑥 + 9 at the

number 𝑎.
Solution: From definition 2.3 we have

′(
𝑓 (𝑥 ) − 𝑓 (𝑎 ) (𝑥 2 − 8𝑥 + 9) − (𝑎2 − 8𝑎 + 9)
𝑓 𝑎) = lim = lim
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎
(𝑥 2−𝑎 2 )−8(𝑥−𝑎)
= lim
𝑥→𝑎 𝑥−𝑎
(𝑥−𝑎)(𝑥+𝑎)−8(𝑥−𝑎)
= lim
𝑥→𝑎 𝑥−𝑎

= 2𝑎 − 8

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 Activity 2.3

a. Find the derivative with respect to a of 𝒇(𝒙) = 𝒙𝟑 − 𝒙

b. Graph 𝒇 and 𝒇′ together and discuss the relationship between the two
graph.

2.2. DIFFERENTIABLE FUNCTION

The derivative as a function
The function 𝑓 ′ that arises when we take the derivative of 𝑓 at such numbers is
called the derivative of 𝑓 (or the derivative of 𝑓 with respect to 𝑥). Thus 𝑓 ′ is by
definition the function whose domain is the collection of numbers at which 𝑓 is
differentiable and whose value at any such number 𝑥 is given by
(4) 𝑓(𝑡) − 𝑓(𝑥)
𝑓 ′ (𝑥 ) = lim
(4) 𝑡 − 𝑥
𝑡→𝑥

Notice that in (4), 𝑥 represents any number at which 𝑓 is differentiable. However,

when the limit on the right side of (4) is evaluated, 𝑡 is the variable and 𝑥 regardes
as a constant.
Example 2.7. Let (𝑥 ) = 𝑥 2 . Show that 𝑓 ′ (𝑥 ) = 2𝑥 for all 𝑥
Solution: It follows from (4) that for all 𝑥
𝑓(𝑡)−𝑓(𝑥) 𝑡 2 −𝑥 2 (𝑡−𝑥)(𝑡+𝑥)
𝑓 ′ (𝑥 ) = lim = lim = lim = lim(𝑡 + 𝑥) = 2𝑥
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥 𝑡−𝑥 𝑡→𝑥 𝑡−𝑥 𝑡→𝑥

Activity 3.4. Let 𝒇(𝒙) = 𝒙𝟑 . show that 𝒇′ (𝒙) = 𝟑𝒙𝟐

1⁄ 1 −1⁄
Example 2.8. Let 𝑓 (𝑥 ) = 𝑥 2 .show thatf ′ (x) = 2 x 2.

Solution: From (4) we find that for 𝑥 > 0

1 1⁄
′(
𝑓 (𝑡 ) − 𝑓 (𝑥 ) 𝑡 ⁄2 − 𝑥 2
𝑓 𝑥 ) = lim = lim
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥 𝑡−𝑥

4
 1⁄ 1⁄ 1⁄ 1⁄
𝑡 −𝑥
2 2 𝑡 2 +𝑥 2 t−x
lim . 1 1 = lim ( 1⁄ 1⁄ )
𝑡→𝑥 𝑡−𝑥 𝑡 ⁄2 + 𝑥 ⁄2 t→x (t − x)(𝑡 2 +𝑥 2)

1 1 1 1 −1
lim = = = x ⁄2
𝑡→𝑥 𝑡 1⁄2 +
1
𝑥 ⁄2
1
𝑥 ⁄2 +
1
𝑥 ⁄2
1
2x ⁄2 2

Other notations for the derivative

The expression 𝑓 ′ (x) is neither the only notation for the derivative nor the oldest.
Some of the other notations for derivatives are
dy d
𝑢̇ , , f(x)
dx dx
In table 2.1 we list four common ways of expressing derivatives and show the
relationships between them.
Table 2.1
Function Derivative as a function Derivative at a point
𝑓(𝑥) = 𝑥 2 𝑓 ′ (𝑥) = 2𝑥 𝑓 ′ (4)=8
𝑦 = 𝑥2 𝑑𝑦 𝑑 2 𝑑𝑦
= 2𝑥 𝑜𝑟 (𝑥 ) = 2𝑥 | =8
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥=4
𝑓(𝑥) = 𝑥 2 𝑑 𝑑 2 𝑑
(𝑓(𝑥)) = 2𝑥 𝑜𝑟 (𝑥 ) = 2𝑥 (𝑓(𝑥))| =8
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑥=4

𝑢 = 𝑡2 𝑢̇ = 2𝑡 𝑢̇ |𝑡=4 = 8
𝑑𝑦
Example 3.9. Let 𝑓 (𝑥 ) = −5𝑥 + 9. 𝑓𝑖𝑛𝑑 |
𝑑𝑥 𝑥=2

Solution: Put 𝑦 = 𝑓(𝑥 ) = −5𝑥 + 9

𝑑𝑦 𝑑
= (−5𝑥 + 9) = −5
𝑑𝑥 𝑑𝑥
𝑑𝑦
Now | = −5
𝑑𝑥 𝑥=2

𝒅𝒚 𝟏
Activity 2.5 Find 𝒅𝒙| where 𝒚 = 𝟒
𝒙=𝟑

DIFFERENTIABILITY
Recall from Definition 2.2 that the derivative of a function 𝑓 is defined at those points
where the limit (4) exists. Points where this limit exists are called points of

5
differentiability for 𝑓 and points where this limit does not exist are called points of non
differentiability.
If 𝑥0 is a point of differentiability for 𝑓 then we say that 𝑓 is differentiable at 𝑥0 or that
the derivative of 𝑓 exists at 𝑥0 or that the derivative of 𝑓 exists at 𝑥0 .

Definition 2.4.
If f is differentiable at each number in its domain, then f is a
differentiable function.

Example 2.10. Let 𝑓(𝑥 ) = 𝑐, where 𝑐 is a constant. Show that𝑓 ′ (𝑥 ) = 0 for all 𝑥.
Solution: By the definition of the derivative
𝑓(𝑡)−𝑓(𝑥) 𝑐−𝑐
(5) 𝑓 ′ (𝑥) = lim = lim 𝑡−𝑥 = 0, for all 𝑥
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥

Example 2.11. Let 𝑓 (𝑥 ) = 𝑥.show that𝑓 ′ (𝑥 ) = 1, for all 𝑥

Solution: By the definition of the derivative
𝑓(𝑡)−𝑓(𝑥) 𝑡−𝑥
𝑓 ′ (𝑥 ) = lim = lim 𝑡−𝑥 = 1 for all 𝑥
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥

Example 2.12. Let 𝑓(𝑥 ) = 𝑥 𝑛 . Show that 𝑓 ′ (𝑥 ) = 𝑛𝑥 𝑛−1 , for all 𝑥

Solution: By the definition of derivative
𝑓(𝑡) − 𝑓(𝑥) 𝑡𝑛 − 𝑥𝑛
𝑓 ′ (𝑥 ) = lim = lim
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥 𝑡 − 𝑥

But 𝑡 𝑛 − 𝑥 𝑛 = (𝑡 − 𝑥 )(𝑡 𝑛−1 + 𝑡 𝑛−2 𝑥 + 𝑡 𝑛−3 𝑥 2 + ⋯ + 𝑡𝑥 𝑛−2 + 𝑥 𝑛−1 )

(𝑡−𝑥)(𝑡 𝑛−1+𝑡 𝑛−2 𝑥+𝑡 𝑛−3 𝑥 2+⋯+𝑡𝑥 𝑛−2+𝑥 𝑛−1 )
so, 𝑓 ′ (𝑥 ) = lim
𝑡→𝑥 𝑡−𝑥

= lim(𝑡 𝑛−1 + 𝑡 𝑛−2 𝑥 + 𝑡 𝑛−3 𝑥 2 + ⋯ + 𝑡𝑥 𝑛−2 + 𝑥 𝑛−1 )

𝑡→𝑥

= (𝑥 𝑛−1 + 𝑥 𝑛−2 𝑥 + 𝑥 𝑛−3 𝑥 2 + ⋯ + 𝑥 𝑥 𝑛−2 + 𝑥 𝑛−1 )

= (𝑥 𝑛−1 + 𝑥 𝑛−1 + 𝑥 𝑛−1 + ⋯ + 𝑥 𝑛−1 + 𝑥 𝑛−1 ) = 𝑛𝑥 𝑛−1
Therefore, 𝑓 ′ (𝑥 ) = 𝑛𝑥 𝑛−1 , for all 𝑥
Note: In the Leibniz notation the formula is
𝑑
(6) (𝑥 𝑛 ) = 𝑛𝑥 𝑛−1
𝑑𝑥

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Recall from section 2.2 that 𝑓 ′ (𝑎) is sometimes called the rate of change of 𝑓 at 𝑎.
If the function is given by expressing a variable 𝑦 in terms of a variable 𝑥, we call
𝑑𝑦
the derivative 𝑑𝑥 the rate of change of 𝑦 with respect to 𝑥.

Example 2.13. Find the rate of change of the volume of a cube with respect to its
side length.
Solution: Let 𝑉 be the volume of a cube and 𝑆 be the length of a cube.
Then 𝑉 = 𝑆 3
𝑑𝑉 𝑑𝑉
We are to find 𝑑𝑆 . From (6) with 𝑛 = 3, it follows that 𝑑𝑆 = 3𝑆 2

Thus, the rate of change of the volume with respect to the side length is 3𝑆 2 . That
is 3 times the area of a side.

Activity 2.6 Find the rate of change of the volume of a sphere with respect
to the radius.

Example 2.14. Let 𝑓 (𝑥 ) = sin 𝑥 find 𝑓 ′ (𝑥).

Solution: using the alternative formula for computing the derivative, we have
𝑓(𝑥+ℎ)−𝑓(𝑥) sin(𝑥+ℎ)−sin 𝑥
𝑓 ′ (𝑥 ) = lim = lim
ℎ→0 ℎ ℎ→0 ℎ
sin 𝑥 cos ℎ+sin ℎ cos 𝑥−sin 𝑥
= lim
ℎ→0 ℎ
sin 𝑥(cos ℎ−1) sin ℎ cos 𝑥
= lim ( + )
ℎ→0 ℎ ℎ
sin 𝑥(cos ℎ−1) sin ℎ cos 𝑥
= lim + lim
ℎ→0 ℎ ℎ→0 ℎ
cos ℎ−1 sin ℎ
= sin 𝑥 lim + cos 𝑥 lim
ℎ→0 ℎ ℎ→0 ℎ
cos ℎ−1 sin ℎ
But we have, lim = 0 and lim =1
ℎ→0 ℎ ℎ→0 ℎ

So,
cos ℎ − 1 sin ℎ
𝑓 ′ (𝑥 ) = sin 𝑥 lim + cos 𝑥 lim = sin 𝑥(0) + cos 𝑥 (1) = cos 𝑥
ℎ→0 ℎ ℎ→0 ℎ

Consequently,
𝑑
(sin 𝑥 ) = cos 𝑥 for all 𝑥.
𝑑𝑥

Activity 2.7 Let 𝒇(𝒙) = 𝒄𝒐𝒔 𝒙, find 𝒇′ (𝒙)

7
The function 𝑐, 𝑥 𝑛 , sin x and cos x, which we have discussed in this section, are
differentiable. However, there are functions that are not differentiable.
Example 2.15. Let 𝑓 (𝑥 ) = |𝑥|. Show that f is not differentiable.
𝑥 𝑖𝑓 𝑥 ≥ 0
Solution: 𝑓 (𝑥 ) = |𝑥| = {
−𝑥 𝑖𝑓 𝑥 < 0
If 𝑥 > 0, then 𝑓 (𝑥 ) = 𝑥 and whenever 𝑡 is close enough to 𝑥, we have 𝑡 > 0and
thus 𝑓(𝑡) = 𝑡. therefore,
𝑓(𝑡)−𝑓(𝑥) 𝑡−𝑥
𝑓 ′ (𝑥 ) = lim = lim 𝑡−𝑥 = 1 for all 𝑥 > 0
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥

If 𝑥 < 0, then 𝑓 (𝑥 ) = −𝑥. Whenever 𝑡 is closed enough to 𝑥, we have 𝑡 < 0 and

thus 𝑓(𝑡) = −𝑡 .Therefore,
𝑓(𝑡)−𝑓(𝑥) −𝑡−(−𝑥)
𝑓 ′ (𝑥 ) = lim = lim = −1 for all 𝑥 < 0
𝑡→𝑥 𝑡−𝑥 𝑡→𝑥 𝑡−𝑥

Therefore, f does not have a derivative at 0, and consequently f is not differentiable

function.
Theorem 2.1. If 𝑓 is differentiable at 𝑎, then 𝑓 is continuous at 𝑎.
Proof: Suppose 𝑓 is differentiable at 𝑎, that is
𝑓(𝑥)−𝑓(𝑎)
𝑓 ′ (𝑎) = lim exist.
𝑥→𝑎 𝑥−𝑎

We went to show that f is continuous at a, we have to show that lim 𝑓 (𝑥 ) = 𝑓(𝑎) .

𝑥→𝑎

we do this by showing that the difference 𝑓 (𝑥 ) − 𝑓(𝑎) approaches to 0.

𝑓 (𝑥) − 𝑓(𝑎)
𝑓 (𝑥 ) − 𝑓 (𝑎 ) = (𝑥 − 𝑎)(𝑥 ≠ 𝑎)
𝑥−𝑎
Thus, using the product law and (3), we can write
𝑓(𝑥 ) − 𝑓(𝑎)
lim [𝑓(𝑥 ) − 𝑓(𝑎)] = lim (𝑥 − 𝑎)
𝑥→𝑎 𝑥→𝑎 𝑥−𝑎
𝑓(𝑥 ) − 𝑓(𝑎)
= lim lim (𝑥 − 𝑎)
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎

= 𝑓 ′ ( 𝑎 ). 0 = 0
To use what we have just proved, we start with 𝑓(𝑥) and add and subtract 𝑓(𝑎)
lim 𝑓(𝑥) = lim [𝑓(𝑥) − 𝑓(𝑎) + 𝑓(𝑎)] = lim 𝑓(𝑥) − lim 𝑓(𝑎) + lim 𝑓(𝑎)
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎

= 𝑓 (𝑎 ) − 𝑓 (𝑎 ) + 𝑓 (𝑎 )
= 𝑓(𝑎)
Therefore, f is continuous at a.

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Note: The converse of Theorem 2.1 is false, that is , there are functions that are
continuous but not differentiable.
Question: Give an example show that the converse of theorem is not true.

DIFFERENTIABILITY ON INTERVALS
If 𝑓 is differentiable at every point in an open interval (𝑎, 𝑏) then we say that f is
differentiable on (𝑎, 𝑏). This definition also applies to infinite open intervals of the
form (𝑎, ∞), (−∞, 𝑎)and (−∞, ∞).ln the case where 𝑓 is differentiable on (−∞, ∞)
we say f that is differentiable everywhere.
We say that 𝑓 is differentiable on [𝑎, 𝑏] if f is differentiable on (𝑎, 𝑏) and if the
one-sided limits
𝑓(𝑥)−𝑓(𝑎) 𝑓(𝑥)−𝑓(𝑏)
lim+ and lim−
𝑥→𝑎 𝑥−𝑎 𝑥→𝑏 𝑥−𝑏
(7)

both exist. Of course, if 𝑓 is differentiable at 𝑎 or at 𝑏 then the corresponding one-

sided limit in (5) exists. However, there are functions for which the one-sided limit
in (5) exist even though 𝑓 is not differentiable at 𝑎 or at 𝑏. the absolute function is
an example of such a function.
1⁄
Example 2.16. Let 𝑓(𝑥) = 𝑥 2 . Show that 𝑓 is differentiable on (0, ∞) but not on
[0, ∞)
1 −1⁄
Solution: By equation (6), 𝑓 ′ (𝑥 ) = 2 𝑥 2 for 𝑥 > 0

Consequently, f is differentiable on (0, ∞). For 𝑥 = 0, we have

1
𝑓 (𝑡 ) − 𝑓 (0) 𝑡 ⁄2 −1
lim+ = lim+ = 𝑡 ⁄2 = ∞
𝑡→0 𝑡−0 𝑡→0 𝑡
Since the right-hand sided limit at o doesn’t exist, we conclude that 𝑓 is not
differentiable on [0, ∞).

𝟏
Activity 2.8 Let 𝒇(𝒙) = 𝟒−𝒙𝟐 , show that 𝒇 if differentiable on (−∞, 𝟐)

9
Exercise 2.2.

1. Use the result of this section to find the derivative of the given function at
the given numbers
a. f(x) = −2: a = 1
1
b. f(x) = x 4 : a = − 2
3
c. f(x) = |x|, : a = −11,
4
π π
d. f(x) = sin x ; a = 4 , 3

2. find the derivative of the function

a. y = √x − 1
b. y = cos 2x
3. Show that f is differentiable on the given interval
a. f(x) = 2x 2 − √x ; (0, ∞)f(x) = |x − 1| ; [1, ∞)
b. f(x) = x 3 − x; (−∞, ∞)
1
c. f(x) = ; (−∞, 2)
4−x2

2.3 DERIVATIVES OF COMBINATIONS OF FUNCTIONS.

Objective: On completion of this section the students should be able to:

✓ Understand the derivative of sums, differences, products

and quotients of functions,
✓ Using the rule of differentiation find the derivative of a
function

Since there are limit theorems for sums, differences, products and quotients of
functions, it is natural to ask whether there are corresponding theorems for
derivatives, there are such theorems, but some of the formulas for the derivatives
of combinations of functions are quite different from their counter parts for limits.

10
Theorems 2.2. If 𝑓 and 𝑔 are differentiable at 𝑎, and 𝑐 be any constant number,
then

a. The function 𝑓 + 𝑔 is differentiable and (𝑓 + 𝑔)′ (𝑎) = 𝑓 ′ (𝑎) + 𝑔′ (𝑎)

b. 𝑐𝑓is differentiable and (𝑐𝑓)′ (𝑎) = 𝑐𝑓 ′ (𝑎)
c. 𝑓 − 𝑔 is differentiable and (𝑓 − 𝑔)′ (𝑎) = 𝑓 ′ (𝑎) − 𝑔′ (𝑎)
d. 𝑓𝑔 is differentiable and (𝑓𝑔)′ (𝑎) = 𝑓 ′ (𝑎)𝑔(𝑎) + 𝑓(𝑎)𝑔′ (𝑎)
𝑓 𝑓 ′ 𝑓′ (𝑎)𝑔(𝑎)−𝑓(𝑎)𝑔′(𝑎)
e. is differentiable and (𝑔) (𝑎) = [𝑔(𝑎)]2
, 𝑔 (𝑎 ) ≠ 0
𝑔

Proof: a) using the limit theorems, we find that

(𝑓 + 𝑔)(𝑥 ) − (𝑓 + 𝑔)(𝑎)
(𝑓 + 𝑔)′ (𝑎) = lim
𝑥→𝑎 𝑥−𝑎

(𝑓(𝑥)+𝑔(𝑥))−(𝑓(𝑎)+𝑔(𝑎))
= lim
𝑥→𝑎 𝑥−𝑎

𝑓(𝑥)−𝑓(𝑎) 𝑔(𝑥)−𝑔(𝑎)
= lim ( + )
𝑥→𝑎 𝑥−𝑎 𝑥−𝑎
𝑓(𝑥)−𝑓(𝑎) 𝑔(𝑥)−𝑔(𝑎)
= lim + lim
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎

= 𝑓 ′ (𝑎) + 𝑔′ (𝑎)
Therefore, (𝑓 + 𝑔)′ (𝑎) = 𝑓 ′ (𝑎) + 𝑔′ (𝑎)

b). From the definition of differentiability at 𝑎, we have

(𝑐𝑓)(𝑥 ) − (𝑐𝑓)(𝑎) 𝑐𝑓 (𝑥 ) − 𝑐𝑓(𝑎)
(𝑐𝑓)′ (𝑎) = lim = lim
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎
𝑐(𝑓(𝑥) − 𝑓(𝑎))
= lim
𝑥→𝑎 𝑥−𝑎
𝑓 (𝑥 ) − 𝑓(𝑎)
= 𝑐 lim = 𝑐 𝑓 ′ (𝑎 )
𝑥→𝑎 𝑥−𝑎
Therefore, (𝑐𝑓)′ (𝑎) = 𝑐𝑓 ′ (𝑎)
c) exercise
d) By the definition of differentiation at 𝑎 and applying several limit theorems,
we have
(𝑓𝑔 )(𝑥 ) − (𝑓𝑔)(𝑎) 𝑓 (𝑥 )𝑔(𝑥 ) − 𝑓 (𝑎)𝑔(𝑎)
(𝑓𝑔)′ (𝑎) = lim = lim
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎

11
 𝑓(𝑥 )𝑔(𝑥 ) − 𝑓 (𝑎)𝑔(𝑥 ) + 𝑓(𝑎)𝑔(𝑥 ) − 𝑓 (𝑎)𝑔(𝑎)
= lim
𝑥→𝑎 𝑥−𝑎
𝑓(𝑥 )𝑔(𝑥 ) − 𝑓(𝑎)𝑔(𝑥 ) 𝑓(𝑎)𝑔(𝑥 ) − 𝑓 (𝑎)𝑔(𝑎
= lim ( + )
𝑥→𝑎 𝑥−𝑎 𝑥−𝑎
𝑔(𝑥 )[𝑓(𝑥 ) − 𝑓 (𝑎)] 𝑓(𝑎)[𝑔(𝑥 ) − 𝑔(𝑎)]
= lim + lim
𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥−𝑎
𝑓(𝑥) − 𝑓(𝑎) 𝑔(𝑥 ) − 𝑔(𝑎)
= lim 𝑔(𝑥) lim + lim 𝑓(𝑎) lim
𝑥→𝑎 𝑥→𝑎 𝑥−𝑎 𝑥→𝑎 𝑥→𝑎 𝑥−𝑎
= 𝑔(𝑎)𝑓 ′ (𝑎) + 𝑓(𝑎)𝑔′ (𝑎)
Therefore, (𝑓𝑔)′ (𝑎) = 𝑓 ′ (𝑎)𝑔(𝑎) + 𝑓(𝑎)𝑔′ (𝑎)
e) Exercise

Example 2.17. Differentiate the following

a) 𝑓 (𝑥 ) = 3√𝑥
𝑑 𝑑 1⁄
Solution: 𝑓 ′ (𝑥 ) = 𝑑𝑥 ( 3√𝑥 ) = 𝑑𝑥 (𝑥 3)

1 −2⁄
Using (6), 𝑓 ′ (𝑥) = 3 𝑥 3

𝑑 3
b) (3𝑥 8 − 12𝑥 5 − √2𝑥 4 − 2 𝑥 3 − 6𝑥 + 5) and evaluate it at 𝑥 = 1.
𝑑𝑥

Solution: first using addition and subtraction rule to find the derivative of the
function at an arbitrary 𝑥
𝑑 3
(3𝑥 8 − 12𝑥 5 − √2𝑥 4 − 𝑥 3 − 6𝑥 + 5)
𝑑𝑥 2
3
= 3(8𝑥 7 − 12(5𝑥 4 )) − √2(4𝑥 3 ) − (3𝑥 2 ) − 6
2
9
= 24𝑥 7 − 60𝑥 4 − 4√2𝑥 3 − 𝑥 2 − 6
2
9
So, 𝑓 ′ (𝑥) = 24𝑥 7 − 60𝑥 4 − 4√2𝑥 3 − 2 𝑥 2 − 6

Then we evaluate the derivative at 𝑥 = 1

9 93
𝑓 ′ (1) = 24(1)7 − 60(1)4 − 4√2(1)3 − (1)2 − 6 = − − 4√2
2 2
c) 𝑓 (𝑥 ) = 𝑥 2 cos 𝑥
Solution: Let ℎ(𝑥) = 𝑥 2 and 𝑔(𝑥 ) = cos 𝑥so, 𝑓 (𝑥 ) = ℎ(𝑥)𝑔(𝑥)and
ℎ′ (𝑥 ) = 2𝑥, 𝑔′ (𝑥 ) = − sin 𝑥
Using theorem 3.2 (d),

12
 𝑓 ′ (𝑥 ) = ℎ′ (𝑥 )𝑔(𝑥 ) + ℎ(𝑥 )𝑔′ (𝑥 ) = 2𝑥(cos 𝑥 ) + 𝑥 2 (− sin 𝑥)
= 2𝑥 cos 𝑥 − 𝑥 2 sin 𝑥
d) 𝑓(𝑥 ) = √𝑥𝑔(𝑥 ), where 𝑔(4) = 2 and𝑔′ (4) = 3. Find 𝑓 ′ (4)
Solution: Applying product rule, we get
′
𝑓 ′ (𝑥 ) = (√𝑥 𝑔(𝑥))′ = (√𝑥) 𝑔(𝑥) + √𝑥𝑔′ (𝑥)
1⁄ ′
= (𝑥 2 ) 𝑔 (𝑥 ) + √𝑥𝑔′ (𝑥)
1 −1 ′
= 𝑥 ⁄2𝑔(𝑥)+√𝑥𝑔 (𝑥))
2
1 −1⁄ 1 13
So, 𝑓 ′ (4) = 2 (4) 2 𝑔 (4) + √4𝑔′ (4) = 4 . 2 + 2.3 = 2

e) 𝑓(𝑥 ) = sec 𝑥
1
Solution: 𝑓 (𝑥 ) = sec 𝑥 = cos 𝑥

Applying quotient rule, we get

𝑑 𝑑
𝑑 1 (1) cos 𝑥 − 1. cos 𝑥 −(− sin 𝑥) sin 𝑥
𝑓 ′ (𝑥 ) = ( ) = 𝑑𝑥 2
𝑑𝑥
= 2
=
𝑑𝑥 cos 𝑥 (cos 𝑥 ) 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 2 𝑥
sin 𝑥 1
=
cos 𝑥 cos 𝑥
= tan 𝑥 sec 𝑥
𝑑
So,𝑓 ′ (𝑥) = 𝑑𝑥 (sec 𝑥 ) = tan 𝑥 sec 𝑥

Activity2 .9.
a) differentiate the function 𝒇(𝒕) = √𝒂(𝒂 + 𝒃𝒕) ,where 𝒂 and 𝒃 are
constant.
𝒅
b) Show that 𝒅𝒙 (𝒄𝒔𝒄 𝒙) = − 𝒄𝒔𝒄 𝒙 𝒄𝒐𝒕 𝒙

Exercise 2.3.

1. Find the derivative of the given function.

x2 +1
a) f(x) = x3 +1 d) f(x) = −4x 3
A
b) f(x) = √xex e) y =
B+Cex

13
 1 3
c) y = (x2 − x4 ) (x + 5x 2 ) f) y = √x sec x

2. Let f(x) = |x| and g(x) = −|x|. find a simple formula for f + g .
3. Suppose that f ′ (a), g ′ (a)andh′ (a) exist. Prove that
a) (f + g + h)′ (a) = f ′ (a) + g ′ (a) + h′ (a)
b) (fgh)′ (a) = f ′ (a)g(a)h(a) + f(a)g ′ (a)h(a) + f(a)g(a)h′ (a)
d
c) Taking f = g = h in part (b). show that dx [f(x)]3 = 3(f(x))2 f ′ (x)

4. Suppose that f(5) = 1, f ′ (5) = 6, g(5) = −3 andg ′ (5) = 2. Find

a) (fg)′ (5) d) (f + g), (5)
f ′ f ′
b) (g) (5) e) (f−g) (5)

g ′
c) ( ) (5)
f

5. a) If f(x) = ex g(x), where g(0) = 2 andg ′ (0) = 5 findf ′ (0)

d f(x)
b) If f(x) =4 and f ′ (2) = −3 find dx ( ) at x = 2
x

2.4 THE CHAIN RULE

Objective: On completion of this section the students should be able to:
✓ Understand the chain rule
✓ Using the rule solve the questions
Suppose you are asked to differentiate the function
𝐹(𝑥 ) = sin 3𝑥
The differentiation formula you learned in the previous sections of this chapter
can be applied to calculate 𝐹 ′ (𝑥), however it is not simple. Observe that F is a
composite function. In fact, if we let 𝑓(𝑥 ) = 3𝑥 and 𝑔(𝑥 ) = sin 𝑥 and 𝐹 (𝑥 ) =
𝑔°𝑓. We know how to differentiate both 𝑓 and 𝑔, so it would be useful to
have a rule that tells us how to find the derivative of 𝐹 (𝑥 ) = 𝑔𝑜𝑓 in terms of
the derivative of 𝑓 and 𝑔.
The Chain Rule:
If 𝑓 is differentiable at 𝑎 and 𝑔 is differentiable at 𝑓(𝑎). Then 𝑔𝑜𝑓 is
differentiable at 𝑎, and
(𝑔 𝑜𝑓)′ (𝑎) = 𝑔′ (𝑓(𝑎))𝑓 ′ (𝑎)
The chain rule can be written for 𝑥 in the notation as

14
 (𝑔 𝑜𝑓)′ (𝑥 ) = 𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥)

Example 2.18. Let (𝑥) = sin 3𝑥 . Find a formula for 𝐹 ′ (𝑥).

Solution: Let 𝑓(𝑥) = 3𝑥 and 𝑔(𝑥) = 𝑠𝑖𝑛 𝑥. then 𝐹 = 𝑔𝑜𝑓, since

𝑓 ′ (𝑥 ) = 3 and𝑔′ (𝑥 ) = cos 𝑥
We conclude that
𝐹 ′ (𝑥 ) = (𝑔𝑜𝑓)′ (𝑥 ) = 𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥)
= cos(3𝑥 )3
= 3 cos 3𝑥

𝑑
Example2.19. Find a formula for 𝑑𝑥 (√𝑥 2 + 1).

Solution: Let 𝐹 (𝑥 ) = √𝑥 2 + 1, 𝑓(𝑥 ) = √𝑥 and 𝑔(𝑥 ) = 𝑥 2 + 1, so 𝐹(𝑥 ) =

1
𝑓(𝑔(𝑥 )) and 𝑓 ′ (𝑥 ) = 2√𝑥 , 𝑔′ (𝑥 ) = 2

By chain rule

𝑑 𝑑 1
(√𝑥 2 + 1) = (𝑓°𝑔)(𝑥) = 𝑓 ′ (𝑔(𝑥))𝑔′ (𝑥) = 2𝑥
𝑑𝑥 𝑑𝑥 2√𝑥 2 + 1
𝑥
=
2
√𝑥 + 1

Activity 2.10 differentiate

𝒙−𝟐 𝟓
a) 𝒄𝒐𝒔𝟐 𝒙 b) (𝟐𝒙+𝟏)

The chain rule assumes a very suggestive form in the Leibniz notation.
Suppose the function f and g in the chain rule are already given and let
𝑑𝑢 𝑑𝑢
𝑢 = 𝑓(𝑥 ) and 𝑦 = 𝑔(𝑢). Then 𝑦 = 𝑔(𝑓(𝑥 )) , = 𝑓 ′ (𝑥 ) and = 𝑔′ (𝑢).
𝑑𝑥 𝑑𝑥

Therefore
𝑑𝑦 𝑑
= (𝑔(𝑓(𝑥 ))) = 𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥)
𝑑𝑥 𝑑𝑥
= 𝑔′ (𝑢)𝑓 ′ (𝑥)

15
 𝑑𝑦 𝑑𝑢
= 𝑑𝑢 𝑑𝑥

Or more concisely,
𝑑𝑦 𝑑𝑦 𝑑𝑢 (8)
=
𝑑𝑥 𝑑𝑢 𝑑𝑥

If 𝑦 = [𝑔(𝑥)]𝑛 then we can write 𝑦 = 𝑓(𝑢) = 𝑢𝑛 where 𝑢 = 𝑔(𝑥). By using

the chain rule and the power rule we get
𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑𝑢
= = 𝑛𝑢𝑛−1
𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑢
Therefore = 𝑛𝑢𝑛−1 𝑑𝑥 = 𝑛[𝑔(𝑥)𝑛−1 ]𝑔′ (𝑥)
𝑑𝑥

The power rule combined with the chain rule

If n is any real number and 𝑢 = 𝑔(𝑥) is differentiable then
𝑑 𝑛 𝑑𝑢
(𝑢 ) = 𝑛𝑢𝑛−1
𝑑𝑥 𝑑𝑥 (9)
𝑑𝑦
Example 2.20 Let 𝑦 = 𝑠𝑖𝑛8 𝑥 find 𝑑𝑥

Solution: put 𝑢 = sin 𝑥and 𝑦 = 𝑢8

Then from (8), it follows that
𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑 𝑑
= = (𝑢8 ) (sin 𝑥)
𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥
= 8𝑢8 cos 𝑥
= 8𝑠𝑖𝑛7 𝑥 cos 𝑥
1
Example 2.21. Find 𝑓 ′ (𝑥) if 𝑓(𝑥 ) = 3
√𝑥 2+𝑥+1
1 𝑑𝑢 −(2𝑥+1) 𝑑𝑦 1 −2⁄
Solution: Let 𝑢 = 𝑥 2 +𝑥+1and 𝑦 = 3√𝑥 . So, 𝑑𝑥 = (𝑥 2+𝑥+1)2and 𝑑𝑥 = 3 𝑢 3

Then from (8) it follows that

𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑑 1 𝑑
= = ( 2 ) ( 3√𝑥 )
𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑥 + 𝑥 + 1 𝑑𝑥
−(2𝑥 + 1) 1 −2⁄
=( ) 𝑢 3
(𝑥 2 + 𝑥 + 1)2 3
⁄3 −2
1 1 −(2𝑥 + 1)
= ( 2 )
3 𝑥 +𝑥+1 𝑥2 + 𝑥 + 1
1 −4
= − (𝑥 2 + 𝑥 + 1) ⁄3 (2𝑥 + 1)
3
Activity 2.11. Differentiate
a) 𝒚 = 𝟑 𝒄𝒐𝒔 𝒏𝜽 𝒃) 𝒚 = 𝒄𝒐𝒕𝟐 (𝒔𝒊𝒏 𝜽)
16
Example 2.22. Suppose the radius 𝑟 of a ballon varies with respect to time
according to the equation 𝑟 = 1 + 2𝑡. Find the rate of change of the ballon’s
volume with respect to time.
4
Solution: Let 𝑉 be the volume, then 𝑉 = 3 𝜋𝑟 3 , while by assumption 𝑟 = 1 + 2𝑡

Therefore (8) tells that

𝑑𝑉 𝑑𝑣 𝑑𝑟
= 𝑑𝑟 𝑑𝑡 = 4𝜋𝑟 2 2 = 8𝜋𝑟 2 = 8𝜋(1 + 2𝑡)2
𝑑𝑡

The compound chain rule:

Let 𝑘(𝑥 ) = (ℎ𝑜𝑔𝑜𝑓)(𝑥 ) = ℎ(𝑔(𝑓(𝑥 ))) and 𝑓 is differentiable at 𝑥, 𝑔
differentiable at 𝑓(𝑥) and ℎ differentiable at 𝑔(𝑓(𝑥))). Since 𝑘(𝑥 ) =
ℎ((𝑔𝑜𝑓)(𝑥)).
A first application of the chain rule yields:
𝑘 ′ (𝑥 ) = ℎ′ ((𝑔𝑜𝑓)(𝑥)(𝑔𝑜𝑓)′ (𝑥), and also (𝑔𝑜𝑓)′ (𝑥 ) = 𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥).
So, 𝑘 ′ (𝑥) = ℎ′ (𝑔(𝑓(𝑥 )))𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥).
Therefore
𝑘 ′ (𝑥 ) = ℎ′ (𝑔(𝑓 (𝑥)))𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥) (10)

In the formula, the derivative of ℎ at the number 𝑔(𝑓(𝑥)) appears first, then the
derivative of 𝑔 at the number 𝑓(𝑥) and finally the derivative of 𝑓 at the number 𝑥.

Example 2.23. Let 𝑘(𝑥) = 𝑐𝑜𝑠 3 4𝑥. Find 𝑘 ′ (𝑥 )and calculate 𝑘 ′ (𝜋⁄6)
Solution: Let ℎ(𝑥 ) = 𝑥 3 , 𝑔(𝑥 ) = cos 𝑥 and 𝑓(𝑥 ) = 4𝑥
Then 𝑘(𝑥 ) = ℎ(𝑔(𝑓 (𝑥))) and ℎ′ (𝑥 ) = 3𝑥 2 , 𝑔′ (𝑥 ) = − sin 𝑥 , 𝑓 ′ (𝑥 ) = 4
From (9) we have
𝑘 ′ (𝑥 ) = ℎ′ ((𝑔𝑜𝑓)(𝑥 )𝑔′ (𝑓(𝑥))𝑓 ′ (𝑥)
= 3(cos 4𝑥)2 (−sin 4𝑥 ). 4
= −12 𝑐𝑜𝑠 2 4𝑥 sin 4𝑥
In particular
√3 −3√3
𝑘 ′ (𝜋⁄6) = −12 𝑐𝑜𝑠 2 4(𝜋⁄6) sin 4(𝜋⁄6) = −12(− 1⁄2)2 ( )=
2 2

Activity2.12. Find 𝒌′ (𝒙) where 𝒌(𝒙) = 𝟒√𝒔𝒆𝒄(𝒕𝒂𝒏 𝒙 )

17
Exercise 2.4.
1. Find the derivative of the function
7⁄ −7⁄ 3
a. f(x) = x 6 −x 6 d) f(x) = √x + tan x
b. f(x) = √2x 2 + 3x − 1 e) y = a3 + cos 3 x
√x3 −1
c. f(x) = f) g(t) = (6t 2 − 5)3 (t 2 − 2)4
x2
dy
2. Write the composite function in the form f(g(x)). Then find dx
1 −3⁄
a. y = sin 2 x d) y = (sin x − cos x) 2

b. y = e√x e) y = √2x + 1
c. y = √3x + 1
dy
3. Write the composite in the form of h(f(g(x))). Then find dx
1⁄
a. y = (cos (4x)) 2

b. y = cos 2 (3x 6 )
c. y = sin √2x + 1
4. Find an equation of the tangent line to the graph of f at the given point
2
a. f(x) = 1+e−x at the point (0,1)

b. f(x) = x√2 − x 2 , find f ′ (x)

5. If F(x) = f(g(x), where f(−2) = 8, f ′ (−2) = 4, f ′ (5) = 32 andg ′ (5) =
6, g(5) = −2 find F ′ (5)
6. IfF(x) = f(g(x)), wheref(−2) = 8, f ′ (−2) = 4, f ′ (5) = 3 g(5) =
−2 and g ′ (5) =6 findF ′ (5).
7. If g(x) = √4 + 3f(x)where f(1) = 7 andf ′ (1) = 4.findg ′ (1)

2.5. Application of Chain Rule; Related Rates and Implicit

Differentiation
𝑑𝑦
Recall from section 2.1 that if 𝑦 = 𝑓(𝑥), then the derivative can be interpreted
𝑑𝑥

as the rate of change of y with respect to x. In this section we examine some of the
application of this idea to chemistry and physics.
Let’s recall from section of tangents the basic idea behind rates of change. If 𝑥
changes from 𝑥1 to 𝑥2 , then the change in 𝑥 is

18
 ∆𝑥 = 𝑥2 − 𝑥1
And the corresponding change in 𝑦 is
∆𝑦 = 𝑓 (𝑥2 ) − 𝑓(𝑥1 )
The difference quotient
∆𝑦 𝑓 (𝑥2 ) − 𝑓(𝑥1 )
=
∆𝑥 𝑥2 − 𝑥1
is the average rate of change of 𝑦 with respect to 𝑥 over the interval [𝑥1 , 𝑥2 ] and
can be interpreted as the slope of the secant line PQ in figure (). Its limit as ∆𝑥 → 0
is the derivative,𝑓 ′ (𝑥1 ), which can therefore be interpreated as the instantaneous
rate of change of 𝑦 with respect to 𝑥 or the slope of the tangent line at
𝑝(𝑥1 , 𝑓(𝑥1 )). We write the process in the form
𝑑𝑦 ∆𝑦
= lim
𝑑𝑥 ∆𝑥→0 ∆𝑥
Whenever the function 𝑦 = 𝑓(𝑥) has a specific interpretation in any sciences, its
derivative will have a specific interpretation as a rate of change.

Example 2.25. Suppose that the bigger the ballon is, the harder it is to inflate. In
particular, suppose that when the volume V is greater than 10 cubic inches, the
8
ballon is inflated at the rate of cubic inches per minute. How fast is the radius of
𝑉

the ballon increasing when the radius is 2 inches?

𝑑𝑟
Solution: Our aim is to find 𝑑𝑡 at 𝑡 = 𝑡0 where 𝑡0 is the instant at which 𝑟 = 2. On

the other hand (2) implies that

𝑑𝑉 𝑑𝑟
= 4𝜋𝑟 2 𝑑𝑡 (a)
𝑑𝑟

On the other hand, we have assumed that for 𝑉 > 10

𝑑𝑉 8 8 6
=𝑉=4 = 𝜋𝑟 3 (b)
𝑑𝑟 𝜋𝑟 3
3

𝑑𝑉
Equating the expressing given in (a) and (b) for 𝑑𝑟 , we deduce that
𝑑𝑟 6
4𝜋𝑟 2 = 3
𝑑𝑡 𝜋𝑟
Or equivalently,
𝑑𝑟 3
= 2 5
𝑑𝑡 2𝜋 𝑟
for the time 𝑡0 at which 𝑟 = 2 we have

19
 𝑑𝑟 3 3
| = 2 5=
𝑑𝑡 𝑡=𝑡0 2𝜋 𝑟 64𝜋 2
Consequently, when the radius is 2 inches, the radius is increasing at the rate of
3
inches per minute.
64𝜋2

Activity 2.13. The position of a particle is given by the equation

𝑠 = 𝑓(𝑡) = 𝑡 3 − 6𝑡 2 + 9𝑡
Where t is measured in seconds and s in meters.
a. Find the velocity at time t.
b. What is the velocity after 2 s?
c. When is the particle at rest?
d. When is the particle moving forward (that is, in the positive
direction)?

2.6. Higher Order Derivatives

Objective: After completing this section the student will be able to:
✓ Define the higher order derivative of a function
✓ Understand the method of finding the higher order
derivative.

If the derivative 𝑓 ′ of a function 𝑓 is itself differentiable, then the derivative of 𝑓 ′

is denoted by 𝑓 ′′ and is called the second derivative of 𝑓. as long as we have
differentiability, we can continue the process of differentiating derivatives to
obtain third, fourth, fifth and even higher derivatives of 𝑓 . The successive
derivatives of 𝑓 are denoted by

𝑓 ′ , 𝑓 ′′ = (𝑓 ′ )′ , 𝑓 ′′′ = (𝑓 ′′ )′ , 𝑓 (4) = (𝑓 ′′′ )′ , …

These are called the first derivative, the second derivative, the third derivative and
so forth. The notation of a derivative of arbitrary order is

𝑓 (𝑛) the nth derivative of 𝑓

20
 Example 2.26. If 𝑓 (𝑥 ) = 3𝑥 5 − 2𝑥 4 + 𝑥 3 − 4𝑥 2 + 2𝑥 + 4. Find the
successive derivative of.
Solution:
𝑓 ′ (𝑥 ) = 15𝑥 4 − 8𝑥 3 + 3𝑥 2 − 8𝑥 + 2
𝑓 ′′ (𝑥 ) = 60𝑥 3 − 24𝑥 2 + 6𝑥 − 8
𝑓 ′′′ (𝑥 ) = 180𝑥 2 − 48𝑥 + 6
𝑓 (4) (𝑥 ) = 360𝑥 − 48
𝑓 (5) (𝑥) = 360
𝑓 (6) (𝑥 ) = 0
.
.
.
𝑓 (𝑛) (𝑥) = 0 (𝑛 ≥ 6)

Example 2.27. Find a general formula for 𝐹 ′′ (𝑥) if 𝐹 (𝑥 ) = 𝑥𝑓(𝑥) and 𝑓 and 𝑓 ′ (𝑥)
are differentiable at 𝑥.

Solution: using the product rule differentiate 𝐹.

𝐹(𝑥 ) = 𝑥𝑓(𝑥)

⬚
⇒ 𝐹 ′ (𝑥 ) = (𝑥𝑓(𝑥))′ = 𝑓 (𝑥 ) + 𝑥𝑓 ′ (𝑥)

⬚ ′
⇒ 𝐹 ′′ (𝑥 ) = (𝑓 (𝑥 ) + 𝑥𝑓 ′ (𝑥))′ = (𝑓(𝑥)) + (𝑥𝑓 ′ (𝑥))′
= 𝑓 ′ (𝑥) + 𝑓 ′ (𝑥 ) + 𝑥𝑓 ′′ (𝑥)
Therefore,
𝐹 ′′ (𝑥 ) = 2𝑓 ′ (𝑥) + 𝑥𝑓 ′′ (𝑥)

Successive derivatives can also be denoted as follows:

𝑑
𝑓 ′ (𝑥 ) = [𝑓(𝑥)]
𝑑𝑥
𝑑 𝑑 𝑑2
𝑓 ′′ (𝑥) = [ [𝑓(𝑥)]] = 2 [𝑓(𝑥)]
𝑑𝑥 𝑑𝑥 𝑑𝑥

21
 𝑑 𝑑2 𝑑3
𝑓 ′′′ (𝑥 ) = [ 2 [𝑓(𝑥)]] = 3 [𝑓(𝑥)]
𝑑𝑥 𝑑𝑥 𝑑𝑥
.
.
.
In general, we write
𝑑𝑛
𝑓 (𝑛) (𝑥) = [𝑓(𝑥)]
(11) 𝑑𝑥 𝑛

Which is read "the nth derivative of 𝑓 with respect to 𝑥. "

When a dependent variable is involved, say y = f(x). Then successive derivatives
can be denoted by writing
𝑑𝑦 𝑑 2 𝑦 𝑑 3 𝑦 𝑑 4 𝑦 𝑑𝑛 𝑦
, 2, 3 , 4 ,…, 𝑛 ,…
𝑑𝑥 𝑑𝑦 𝑑𝑦 𝑑𝑦 𝑑𝑦
or more briefly
𝑦 ′ , 𝑦 ′′ , 𝑦 ′′′ , … , 𝑦 (𝑛) , …

Example 2.28. Find the 27th derivative of 𝑐𝑜𝑠𝑦

Solution: the first few derivatives of 𝑓(𝑦) = 𝑐𝑜𝑠 𝑦 are as follows:
𝑓 ′ (𝑦) = − sin 𝑦
𝑓 ′′ (𝑦) = − cos 𝑦
𝑓 ′′′ (𝑦) = sin 𝑦
𝑓 (4) (𝑦) = cos 𝑦
𝑓 (5) (𝑦) = − sin 𝑦 = 𝑓 ′ (𝑦)
.
.
.
We see that the successive derivatives occur in a cycle of length 4 and, in
particular 𝑓 (𝑛) (𝑦) = cos 𝑦 whenever n is a multiple of 4. Therefore
𝑓 (24) (𝑦) = cos 𝑦
and, differentiating three more times, we have
𝑓 (27) (𝑦) = sin 𝑦

Activity 2.14 Find

a) 𝑦 ′′′ (0), where 𝑦 = 4𝑥 4 + 2𝑥 3 + 3
𝑑4 𝑦 6
b) | , where 𝑦 = 𝑥 4
𝑑𝑥 4 𝑥=1

22
 Exercise 2.6
1. Show that y = x 3 + 3x + 1 satisfiesy ′′′ + xy ′′ + y ′ = 0.
1
2. Show that if x ≠ 0, then y = x satisfies the equation

x 3 y ′′′ + x 2 y ′′′ = xy = 0

3. find f ′ (x), f ′′ (x)and f ′′′ (x)

1
a) f(x) = x −3 + x5

b) f(x) = (3x 2 + 6)(2x − 1)

c) f(x) = (x 3 + 2x 2 − 6x + 8)(2x −1 + x −2 )
d) f(x) = (3x 2 + 4x + 1)2
4. a) Use the Quotient Rule to differentiate the function
tan x−1
f(x) = sec x

b) Simplify the expression for f(x) by writing it in terms of sin x and

cos x and then findf ′ (x)
c) show that your answers to part (a) and (b) are equivalent.
5. Suppose f(π⁄3) = 4 andf ′ (π⁄3) = −2 and let
g(x) = f(x) sin x
cos x
and h(x) = f(x)

Find a) g ′ (π⁄3)and b) h′ (π⁄3).

2.7. Implicit Differentiation

Objective: On completion of this section the students should be able to:
✓ Understand the implicit differentiation
✓ Differentiate the functions using implicit differentiation

23
Up to now, we have been concerned with differentiating functions that are
expressed in the form 𝑦 = 𝑓(𝑥). An equation of this form is said to define 𝑦
explicitly as a function of 𝑥.Because the variable y appears alone on one side of the
equation. However, sometimes functions are defined by equations in which y is not
alone on one side: for example, the equation

𝑦𝑥 + 𝑦 + 1 = 𝑥 (12)
is not of the form 𝑦 = 𝑓(𝑥) . However, this equation still defined y as a function of
x since it can be rewritten as
𝑥−1
𝑦 = 𝑥+1

Thus, we say that (12) defines y implicitly as a function of x, the function being
𝑥−1
𝑓 (𝑥 ) =
𝑥+1
The method of implicit differentiation consists of differentating both sides of the
equation with respect to 𝑥 and then solving the resulting equation for 𝑦 ′ . In the
examples and exercises of this section it is always assumed that the given equation
determines y implicitly as a differentable function of x so that the method of
implicit differentation can be applied.

Example 2.29

𝑑𝑦
a) If 𝑥 2 + 𝑦 2 = 25 find .
𝑑𝑥

b) Find an equation of the tangent to the circle 𝑥 2 + 𝑦 2 = 25 at the point (3,4)

Solution:

a) Differentiate both sides of the equation𝑥 2 + 𝑦 2 = 25

𝑑 2 𝑑
(𝑥 + 𝑦 2 ) = (25)
𝑑𝑥 𝑑𝑥
𝑑 2 𝑑
(𝑥 ) + (𝑦 2 ) = 0
𝑑𝑥 𝑑𝑥
Remembering that is a function of x and using the chain rule, we have
𝑑 2 𝑑 𝑑𝑦 𝑑𝑦
(𝑦 ) = (𝑦 2 ) = 2𝑦
𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑𝑦
Thus, 2𝑥 + 2𝑦 =0
𝑑𝑥

24
 𝑑𝑦
Now we solve this equation for 𝑑𝑥
𝑑𝑦 𝑦
=−
𝑑𝑥 𝑥
b) At the point (3,4) we have 𝑥 = 3 and 𝑦 = 4, so
𝑑𝑦 3
=−
𝑑 4
An equation of the tangent line to the circle at (3,4) is therefore
3
𝑦 − 4 = − 2 (𝑥 − 3)or 3𝑥 + 4𝑦 = 25

Or b) Solving the equation 𝑥 2 + 𝑦 2 = 25 , we get 𝑦 = ±√25 − 𝑥 2 . The point

(3,4) lies on the upper semicircle 𝑦 = √25 − 𝑥 2 and so we consider the
function 𝑓 (𝑥 ) = √25 − 𝑥 2. Differentiating f using the Chain Rule, we have
1 𝑑
(25 − 𝑥 2 )−1⁄2
𝑓 ′ (𝑥 ) = (25 − 𝑥 2 )
2 𝑑𝑥
1 𝑥
= (25 − 𝑥 2 )−1⁄2 (−2𝑥 ) = −
2 √25 − 𝑥 2

3 3
So, 𝑓 ′ (3) = − √25−32 = − 4

and an equation of the tangent is 3𝑥 + 4𝑦 = 25.

𝑑𝑦
Example 2.30. Use implicit differentiation to find 𝑑𝑥 if 5𝑦 2 + sin 𝑦 = 𝑥 2

Solution:
𝑑 𝑑
(5𝑦 2 + sin 𝑦) = (𝑥 2 ),
𝑑𝑥 𝑑𝑥
𝑑 𝑑
implies 5 𝑑𝑥 (𝑦 2 ) + 𝑑𝑥 (sin 𝑦) = 2𝑥,
𝑑𝑦 𝑑𝑦
implies 5 (2𝑦 ) + (𝑐𝑜𝑠 𝑦) = 2𝑥,
𝑑𝑥 𝑑𝑥
𝑑𝑦 𝑑𝑦
implies 10𝑦 𝑑𝑥 + (𝑐𝑜𝑠 𝑦) 𝑑𝑥 = 2𝑥
𝑑𝑦
Solving for 𝑑𝑥 we obtain
𝑑𝑦 2𝑥
=
𝑑𝑥 10𝑦 + cos 𝑦

25
 Activity 2.15.
𝑑𝑦
a) Use implicit differentiation to find for the equation 𝑥 3 + 𝑦 3 = 3𝑥𝑦
𝑑𝑥

b) Find an equation for the tangent line to the equation𝑥 3 + 𝑦 3 = 3𝑥𝑦 at the
3 3
point (2 , 2).

𝑑2 𝑦
Example 2.31. Use implicit differentiation to find 𝑑𝑥 2 if 4𝑥 2 − 2𝑦 2 = 9

Solution: Differentiating both sides of 4𝑥 2 − 2𝑦 2 = 9 implicitly yields

𝑑𝑦
8𝑥 − 4𝑦 =0
𝑑𝑥
from which we obtain
𝑑𝑦 2𝑥
= (i)
𝑑𝑥 𝑦

Differentiating both sides of (i) implicitly yields

𝑑2 𝑦 (𝑦)(2)−(2𝑥)(𝑑𝑦 ⁄𝑑𝑥 )
= (ii)
𝑑𝑥 2 𝑦2

Substituting (i) into (ii) and simplifying using the original equation. we obtain
𝑑 2 𝑦 2𝑦 − 2𝑥(2𝑥⁄𝑦) 2𝑦 2 − 4𝑥 2 9
= = =
𝑑𝑥 2 𝑦2 𝑦3 𝑦3
𝑑𝑦
Example 2.32. Find 𝑑𝑥 if sin(𝑥 + 𝑦) = 𝑦 2 cos 𝑥

Solution: Differentiating implicitly, we get

𝑑 𝑑
(sin(𝑥 + 𝑦)) = 𝑑𝑥 (𝑦 2 cos 𝑥),
𝑑𝑥
𝑑 𝑑 𝑑
implies cos(𝑥 + 𝑦) 𝑑𝑥 (𝑥 + 𝑦) = 𝑦 2 𝑑𝑥 (cos 𝑥 ) + (cos 𝑥 ) 𝑑𝑥 (𝑦 2 )
𝑑𝑦 𝑑𝑦
implies cos(𝑥 + 𝑦)(1 + 𝑑𝑥 ) = 𝑦 2 sin 𝑥 + cos 𝑥 (2𝑦 𝑑𝑥 ),
𝑑𝑦 𝑑𝑦
implies cos(𝑥 + 𝑦) + 𝑐𝑜𝑠(𝑥 + 𝑦) 𝑑𝑥 = 𝑦 2 sin 𝑥 + 2ycos 𝑥 ,
𝑑𝑥
𝑑𝑦
implies cos(𝑥 + 𝑦) + 𝑦 2 sin 𝑥 = (2ycos 𝑥 − 𝑐𝑜𝑠(𝑥 + 𝑦)) 𝑑𝑥
𝑑𝑦 𝑐𝑜𝑠(𝑥+𝑦)+𝑦 2 sin 𝑥
implies = 2ycos 𝑥 −𝑐𝑜𝑠(𝑥+𝑦)
𝑑𝑥

26
Exercise 2.7.
dy
1. Find dx

a) xy + 2x + 3x 2 = 4 d) 4x 2 + 9y 2 = 36
1 1
b) +y= 1 e) cos x + √y = 5
x

3 x2 +1
c) y = √4x − 5 f) y = √x2 −5

dy
2. Find dx by implicit differentiation.

a) x 2 + y 2 = 68 f) x 2 y + 3xy 3 − x = 3
x⁄
b) x 3 y 2 − 5x 2 y + x = 1 g) e y =x−y
x+y
c) x 2 = x−y h) ysin(x 2 ) = xsin (y 2 )
y
d) √xy = 1 + x 2 y i) √x + y = 1+x2

e) ey cosx = 1 + sin (xy)

3. Find the slope of the tangent line to the curve at the given points in two
ways: first by solving for y interns of x and differentiating and then by
implicit differentiation.
1 1 1 1
a) x 2 + y 2 = 1: ( , ),( ,− )
√2 √2 √2 √2

b) y 2 − x + 1 = 0: (10,3), (10, −3)

4. Find an equation of the tangent line to the hyperbola
x2 y2
− =1
a2 b 2
at the point (x0 , y0 )
5. Find the value of a and b for the curve x 2 y + ay 2 = b if the point (1,1) is
on its graph and the tangent line at (1,1) has the equation 4x + 3y = 7.

Summery
❖ The function 𝑓 is said to have a derivative at point 𝑎 if
𝑓(𝑥)−𝑓(𝑎) 𝑓(𝑎+ℎ)−𝑓(𝑎)
lim or lim exists.
𝑥→𝑎 𝑥−𝑎 ℎ→𝑜 ℎ

❖ The slope of a function means the derivative of a function.

❖ The equation of a tangent line to the curve 𝑦 = 𝑓(𝑥) at a point
(𝑎, 𝑓(𝑎)) is 𝑦 = 𝑚(𝑥 − 𝑎) + 𝑓(𝑎) where 𝑚 is the slop.

27
 ❖ If the derivative of a function 𝑓 is exist at each point in its domain,
then we say that a function 𝑓 is differentiable.
❖ If 𝑓 is differentiable at 𝑎 then f is contionous.
❖ A function 𝑓 is differentiable on [𝑎, 𝑏] if it is differentiable on
(𝑎, 𝑏) and if the left and right sided limit are existed.
❖ Differentiation rule: if 𝑓 and 𝑔 are differentiable function and c be
any constant number then:
𝑑 𝑑 𝑑
▪ (𝑓 + 𝑔 ) = (𝑓 ) + (𝑔)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑
▪ (𝑐𝑓) = 𝑑 (𝑓)
𝑑𝑥 𝑑𝑥
𝑑 𝑑 𝑑
▪ (𝑓 − 𝑔 ) = (𝑓 ) − (𝑔)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑 𝑑
𝑑 𝑓 (𝑓)𝑔−𝑓 (𝑔)
▪ ( ) = 𝑑𝑥 𝑑𝑥
𝑑𝑥 𝑔 𝑔2
𝑑 𝑑 𝑑
▪ (𝑓𝑔) = (𝑓 ) 𝑔 + 𝑓 (𝑔)
𝑑𝑥 𝑑𝑥 𝑑𝑥
𝑑
▪ (𝑓𝑜𝑔)(𝑎) = 𝑔′ (𝑓(𝑎))𝑓 ′ (𝑎)
𝑑𝑥

Review Exercise
1. Find the derivative of the following function at x
4⁄ 1
a. f(x) = √x(x 2 − 3) 7 c. f(x) = 3
(4−x2 ) ⁄2

4x2 +2 1
b. f(x) = d. f(t) = t 2 sin t
3x−8
dy
2. Find dx
2
a. y = 4x 3 − √3x + 5x c. y = x 2 tan2 x
x2 −x+1
b. y = 3 sin 2x − √x cos x d. y = x2 +x+1

3. Find the equation of the tangent line at the given point.

1⁄
a. f(x) = 3x 3 − 2x 2 + 4; (1,5) c. f(x) = (x − 2) 7

π
b. f(x) = sin x − 3 cos 2x; ( 6 , −1) d. f(x) =
2 sin x for x < 0
{ 2 (0, 0)
3x + 2x for x ≥ 0
dy
4. Find dx by implicit differentiation

28
 a. y(√x + 1) = x
x2
b. x 2 + y 2 = y2

c. xy = √x + √y
dy
5. Assume that x and y are differentiable function of t. find in terms of
dt
dx
x, y and dt

a. xy = 3 b. y = sin xy 2
6. Let f be differentiable at 0 and let g(x) = f(x 2 ). Show that g ′ (0) = 0
7. What is the equation of a tangent line to the parabola y = x 2 at (−2,4)
8. If f(t) = √4t + 1, find f ′′ (2)
π
9. If g(x) = xsin x, find g ′′ ( 6 )

10. If f(x) = 2x find f n (x)

29