▪ Derivatives of Polynomials and Exponential

Functions
▪ The Product and Quotient Rules
Sections 3.1 & 3.2
Prepared by:
Prof. Wafik Lotfallah
Prof. Maged G. Iskander
Dr. Kamal Tawfik
Lecture 10 Objectives
 Find the derivative of a function using:
 The Power Rule
 The Sum and difference rules
 The Constant Multiple Rule
 The Natural Exponential Derivative Rule
 Use differentiation rules to find the equations of
the tangent and normal lines.
Lecture 10 Objectives (cont.)
 Show that if a function is differentiable at a point,
then it must be continuous at that point.
 Show the Product and Quotient Rules.
 Find the derivative of a function using the Product
and Quotient Rules, as well as the Power Rule, the
Sum and Difference Rules, and the Constant
 Use differentiation rules to solve calculus problems
involving velocity, acceleration, as well as tangent
and normal lines.
Rules for the Derivative
Recall the Definition of the derivative:

Other notations for y = f(x):

We can use this definition to get some rules that could

be used to quickly find the derivative, e.g.:
Example 1
 Find the derivative of
𝑓(𝑥) = 𝜋.
 Answer: 𝑓′(𝑥) = 0.

In general, we have:
Example 2
 Find the derivative of
𝑓(𝑥) = 𝑥
 Answer:
𝑓(𝑥) = 1

which can also be read from the picture:

The Power Rule
 Example 𝟑: Find the derivative of 𝑓(𝑥) = 𝑥 3 .
 Answer: 𝑓′(𝑥) = 3𝑥 2 .
 In general, we have:

Note: The above rule holds, for all n.

However, its proof is easy when n is a positive integer
What about power functions with negative integer exponents?
Example 𝟒
1
▪ Find the derivative of 𝑓(𝑥) = .
𝑥
We can rewrite this function as 𝑓(𝑥) = 𝑥 −1 , then
′ −2
1
𝑓 𝑥 = −1 𝑥 = − 2 .
𝑥

What if the exponent is a fraction?

𝑑
▪ Find ( 𝑥)
𝑑𝑥
𝑑
We can rewrite this equation as (𝑥 1Τ2 ),
then
𝑑𝑥
𝑑 1 Τ2 1 −1Τ2 1
𝑥 = 2𝑥 =
𝑑𝑥 2 𝑥
Parallel and Perpendicular Lines
Recall that:
Example 5
Find the derivative of :
1
𝑓 𝑥 = .
𝑥 𝑥
Also, find the equations of the tangent and the normal lines
to the graph of f at the point (1, 1).
Solution
1 1 1
The derivative of 𝑓 𝑥 = = = = 𝑥 −3Τ2 is
𝑥 𝑥 𝑥 𝑥 1Τ2 𝑥 3Τ2

𝑓 ′ 𝑥 = −32 𝑥 −3Τ2 −1 =
−32 𝑥 −5Τ2 .
So the slope of the tangent line at (1,1) is
𝑚 = 𝑓 ′ 1 = −32 .
Therefore an equation of the tangent line is
𝑦 − 1 = −32 𝑥 − 1 ⇒ 𝑦 = −32 𝑥 + 52
The normal line is perpendicular to the tangent line, so its
−3 2
slope is the negative reciprocal of , that is, . Thus an
2 3
equation of the normal line is
𝑦 − 1 = 23 𝑥 − 1 ⇒ 𝑦 = 23 𝑥 + 13
New Derivatives from Old
Example 6
Find the derivatives of :
𝑎) 𝑓 𝑥 = 𝑥 8 + 12𝑥 5 − 4𝑥 4 + 10𝑥 3 − 6𝑥 + 5
2
1
𝑏) 𝑔(𝑥) = 𝑥 +
𝑥
Note: In (b), we first need to expand the expression.
Solution
𝑎) 𝑓′(𝑥) =
𝑏) First of all, we rewrite the function 𝑔(𝑥) as
2
1 2 1 1
𝑔 𝑥 = 𝑥+ = 𝑥 + 2𝑥 𝑥 + 2
𝑥 𝑥
= 𝑥 2 + 2 + 𝑥 −2 .

Therefore ,
′
𝑑 2
𝑔 𝑥 = 𝑥 + 2 + 𝑥 −2
𝑑𝑥
= 2𝑥 + 0 − 2𝑥 −3
2
= 2𝑥 − .
𝑥3
Example 7
Find the points on the curve 𝑦 = 𝑥 4 − 6𝑥 2 + 4 where the
tangent line is horizontal.
Solution
Horizontal tangents occur where the derivative is zero.
We have

𝑑𝑦
Now equate =0 ⇒ 4𝑥 𝑥 2 − 3 = 0
𝑑𝑥

Hence 𝑥 = 0 or 𝑥 2 − 3 = 0 which gives 𝑥 = ∓ 3

So the given curve has horizontal tangents when
𝑥 = 0, 3 and − 3
The corresponding points are (0,4), ( 3, −5) and (− 3, −5)
The Derivative of 𝒇 𝒙 = 𝒃 𝒙
 Question: What is the derivative of the function
𝑓 𝑥 = 𝑏𝑥

 Caution: It is NOT 𝑥𝑏 𝑥−1

 Using the definition of the derivative, we get:
′
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
𝑓 𝑥 = lim
ℎ→0 ℎ

𝑏 𝑥+ℎ − 𝑏 𝑥 𝑏 ℎ−1
= lim = 𝑏 𝑥 ⋅ lim = 𝑏 𝑥 ⋅ 𝑐,
ℎ→0 ℎ ℎ→0 ℎ
𝑏ℎ −1
where 𝒄 = lim = 𝒇′ 𝟎 . Hence
ℎ→0 ℎ
The Derivative of 𝒇 𝒙 = 𝒆 𝒙

Recall that:
the function 𝑓(𝑥) = 𝑒 𝑥 is the one whose tangent line at
(0,1) has a slope exactly 1. That is, 𝑓 ′ 0 = 1
Hence,

Now If we put 𝑏 = 𝑒 and, therefore, 𝑓′(0) = 1, in the equation

𝑓′(𝑥) = 𝑓 ′ 0 𝑏 𝑥 ,
we obtain the following important differentiation formula:
Example 8
At what point on the curve 𝑦 = 𝑒 𝑥 is the tangent line parallel
to the line 𝑦 = 2𝑥.
Solution
The derivative of 𝑦 = 𝑒 𝑥 is 𝑦 ′ = 𝑒 𝑥 .
So the slope of the tangent line at any point (𝑎, 𝑒 𝑎 ) is given by
𝑚 = 𝑦′|𝑥=𝑎 = 𝑒 𝑎 .
This tangent line will be parallel to the line 𝑦 = 2𝑥 if it has the
same slope 2,

Hence 𝑚 = 𝑒 𝑎 = 2 ⇒ 𝑎 = ln 2.
Therefore the required point is 𝑎, 𝑒 𝑎 = (ln 2, 2).
The Product and Quotient Rules
Section 3.2
Differentiability Vs. Continuity
 We saw that a function 𝑓(𝑥) can be continuous for all values of
𝑥, but may fail to be differentiable for some values of 𝑥.
Example: 𝑓 𝑥 = |𝑥|.
On the other hand, we have:

 Proof Idea:
𝑓 𝑥 − 𝑓(𝑎)
lim 𝑓(𝑥) = lim 𝑥 − 𝑎 + 𝑓(𝑎)
𝑥→𝑎 𝑥→𝑎 𝑥−𝑎
= 𝑓 ′ 𝑎 ⋅ 0 + 𝑓 𝑎 = 𝑓(𝑎)
The Product Rule

In prime notation:
′
𝑓𝑔 = 𝑓𝑔′ + 𝑔𝑓′
Example 1
Find the derivative of : 𝑦 = (𝑥 2 + 𝑥 + 1)(𝑥 2 − 𝑥 + 1) in two
ways.
Solution
According to the Product Rule, we have
𝑑 2 𝑑 2
𝑦′ = 𝑥2 +𝑥+1 2
𝑥 −𝑥+1 + 𝑥 −𝑥+1 𝑥 +𝑥+1
𝑑𝑥 𝑑𝑥
= 𝑥 2 + 𝑥 + 1 2𝑥 − 1 + 𝑥 2 − 𝑥 + 1 2𝑥 + 1
= 4𝑥 3 + 2𝑥 (Simplify it yourself)
By performing the multiplication first, we get
𝑦 = (𝑥 2 + 𝑥 + 1)(𝑥 2 − 𝑥 + 1)
= 𝑥4 − 𝑥3 + 𝑥2 + 𝑥3 − 𝑥2 + 𝑥 + 𝑥2 − 𝑥 + 1
𝑦 = 𝑥 4 + 𝑥 2 + 1 ⇒ 𝑦 ′ = 4𝑥 3 + 2𝑥.
Example 2
(𝑎) If 𝑓(𝑥) = 𝑥𝑒 𝑥 , find 𝑓 ′ 𝑥 .
(𝑏) Find the 𝑛𝑡ℎ derivative, 𝑓 𝑛 (𝑥).
Solution
(𝑎) By the Product Rule, we have
(b) Using the Product Rule a second time, we get

Further applications of the Product Rule give

𝑓 ′′′ 𝑥 = (𝑥 + 3)𝑒 𝑥 and 𝑓 4 (𝑥) = (𝑥 + 4)𝑒 𝑥
Hence,
𝑓 𝑛 (𝑥) = (𝑥 + 𝑛)𝑒 𝑥
Example 3
If 𝑓 𝑥 = 𝑥 𝑔(𝑥), where 𝑔(4) = 2 and 𝑔′(4) = 3, find 𝑓′(4).
Solution
Applying the Product Rule, we get
The Quotient Rule

In prime notation:
′
𝑓 𝑔𝑓 ′ − 𝑓𝑔′
=
𝑔 𝑔2
Example 4
𝑒 𝑥 −1
Find the derivative of the function: 𝑦 = .
𝑥
Solution
According to the Quotient Rule, we have
𝑑 𝑥 𝑥 𝑑
𝑑𝑦 𝑥 𝑑𝑥 𝑒 − 1 − (𝑒 − 1) 𝑑𝑥 (𝑥)
=
𝑑𝑥 (𝑥)2

𝑥(𝑒 𝑥 ) − 𝑒 𝑥 − 1 . 1
=
𝑥2
So,
𝑑𝑦 𝑥𝑒 𝑥 − 𝑒 𝑥 + 1
= .
𝑑𝑥 𝑥2
Example 5
𝑒𝑥
Where does the graph of the function 𝑔(𝑥) = have a
𝑥3
horizontal tangent line?
Solution
Recall that: the tangent line to the function 𝑔(𝑥) is
horizontal at the point(s) where 𝑔′ 𝑥 = 0. That is, its slope
𝑚 = 0.
Now, using Quotient Rule, we get:
3 𝑑 𝑥 𝑥 𝑑 3
𝑥 𝑒 −𝑒 (𝑥 )
𝑔′(𝑥) = 𝑑𝑥 𝑑𝑥
(𝑥 3 )2

𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥
=
𝑥6
Now, equate 𝑔′(𝑥) to zero, we obtain,

𝑥 3 𝑒 𝑥 − 3𝑥 2 𝑒 𝑥 3 𝑥 2 𝑥
6
= 0 ⇒ 𝑥 𝑒 − 3𝑥 𝑒 =0
𝑥
Equivalently, 𝑥 2𝑒 𝑥 𝑥 − 3 = 0
Since 𝑒 𝑥 ≠ 0 for all real numbers , moreover 𝑥 can never be
zero as the domain of 𝑔(𝑥) is ℝ − {0}.
Consequently, we have
𝑥 − 3 = 0 which gives 𝑥 = 3.
𝑒𝑥
Therefore, the graph of the function 𝑔(𝑥) = has a
𝑥3
𝑒3
horizontal tangent line at the point (3, 𝑔(3)) = (3, ).
27
Summary of Differentiation Rules
Thank you for listening.