In this chapter, we begin our study
of differential calculus.
This is concerned with how one quantity changes in
relation to another quantity.
the derivative is a measure of how fast does a function
change in response to changes in independent variable;
for example, the derivative of the position of a moving
object with respect to time is the object's velocity.
Definition of INFINITESIMAL
1: taking on values arbitrarily close to but greater than zero
2: immeasurably or incalculably small
<an infinitesimal difference>
Now concentrate and observe:
Let’s assume that we know function 𝑓 = 𝑦 𝑥
at any 𝑥 → graph.
Let’s find the slope of the line joining position P(x = a) and some T .
𝑓 𝑎 + ∆𝑥 − 𝑓(𝑎)
𝑠𝑙𝑜𝑝𝑒 =
∆𝑥
T
𝑓(𝑥)
f (x+Δx) - f(x)
P
x
𝑎
𝑎 + ∆𝑥
Let us consider Δ x intervals that are getting smaller and smaller.
Df = f (x+Δx) - f(x)
𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒
𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑎𝑡 𝑃 =
Q
f (x)
R
𝑓 𝑎 + ∆𝑥 − 𝑓(𝑎)
lim
∆𝑥→0
∆𝑥
S
Dx
T
f (x+Δx) - f(x)
P
Dx
𝑎
ALTHOUGH, both, Δ f and Δ x are becoming
infinitesimally small approaching zero, their ratio is
approaching definite value – slope of tangent line at P
Slopesx of straight lines connecting point P and other
points on the path are approaching the slope of the line
tangent at P.
Explain yourself and me what you saw
“The derivative of f with respect to x is …”
Definition:
The derivative of a function 𝑓 at a fixed number 𝑎 is
𝑓′
𝑓 𝑎 + ∆𝑥 − 𝑓(𝑎)
𝑎 = lim
∆𝑥→0
∆𝑥
provided that this limit exists.
Very often you are going to find
that definition in following form:
𝑓′
𝑓 𝑎 + ℎ − 𝑓(𝑎)
𝑎 = lim
ℎ→0
ℎ
Graphical interpretation of mathematical definition of derivative at
point 𝑎 is the slope of the tangent line to the function 𝑓 at point 𝑎 .
Till now, we considered the derivative of a function 𝑓 at a fixed number 𝑎 .
Now, we change our point of view and let the number 𝑎 vary. Let’s assume
that 𝑎 can take any value of 𝑥 on an open interval 𝐼
Definition:
Let y = 𝑓(𝑥) be a function. The derivative of 𝑓 is the function
𝑓 ′ 𝑥 whose value at x is the limit
𝑓′
𝑓 𝑥 + ℎ − 𝑓(𝑥)
𝑥 = lim
ℎ→0
ℎ
provided this limit exists.
If this limit exists for each x in an open interval I, then
we say that f is differentiable on I.
The derivative is the instantaneous rate of change of a
function with respect its variable. This is equivalent to finding
the slope of the tangent line to the function at a point.
There are many ways to write the derivative of
f  x
“f prime x”
y
“y prime”
dy
dx
df
dx
d
f  x
dx
y  f  x
or
“the derivative of f with respect to x”
“d y d x”
or
“the derivative of y with respect to x”
“d f d x”
or
“the derivative of f with respect to x”
“d dx of f of x”
or
“the derivative of f of x with respect to x”
dy
Dy
 lim
dx Dx 0 Dx
Rates of change
• This means that:
– When the derivative is large (and therefore the curve
is steep, as at the point P in the figure), the y-values change rapidly.
– When the derivative is small, the curve is relatively flat and the yvalues change slowly.
y  f  x
4
3
2
1
𝑥
y  f  x
0
3
1
2
3
4
5
6
7
8
9
2
The derivative is defined at the end points
of a function on a closed interval.
1
The derivative is
the slope of the
original function.
𝑥
0
-1
-2
1
2
3
4
5
6
7
8
9
6
5
y  x2  3
4
3
2
1
-3
-2
-1 0
-1
1
x
2
3
y  lim
 x  h
2

 3  x2  3
h 0
h
-2
-3
x 2  2 xh  h 2  x 2
y  lim
h 0
h
6
5
4
3
2
1
-3 -2 -1 0
-1
-2
-3
-4
-5
-6
1 2 3
x
𝑦′
= lim 2𝑥 + ℎ
ℎ→0
y  2 x
0

Differentiability
A function is differentiable if it has a derivative everywhere in its domain.
To be differentiable, a function must be continuous and smooth.
Functions on closed intervals must have one-sided derivatives
defined at the end points.
Derivatives will fail to exist at:
f  x  x
f  x  x
2
3
cusp
corner
f  x  x
3
1, x  0
f  x  
 1, x  0
vertical tangent
A function has
derivative at a point,
if the left derivative is
equal to the right
derivative at that point.
The left slope must be
equal to the right slope.
discontinuity
Find an equation of the tangent line to the hyperbola 𝑦 =
3
at the point (3, 1).
𝑥
The slope of the tangent at (3, 1) is:
𝑚=
𝑓 3+ℎ −𝑓(3)
lim
ℎ
ℎ→0
= lim
ℎ→0
3
−1
3+ℎ
ℎ
−ℎ
ℎ→0 ℎ(3+ℎ)
= lim
= lim
ℎ→0
1
3
−1
3+ℎ
Eq. of the tangent at the point (3, 1) is 𝑦 − 1 = − (𝑥 − 3)
x + 3y – 6 = 0
The hyperbola and its tangent
are shown in the figure
How do you find normal at (3,1) ?
=−
1
3
Equation of the tangent line to a function 𝑦 = 𝑓(𝑥) at the point 𝑎, 𝑓(𝑎)
𝑦 − 𝑓 𝑎 = 𝑓 ′ 𝑎 (𝑥 − 𝑎)
Equation of the normal line to a function 𝑦 = 𝑓(𝑥) at the point 𝑎, 𝑓(𝑎)
1
𝑦−𝑓 𝑎 =− ′
(𝑥 − 𝑎)
𝑓 𝑎
𝐻𝑜𝑤 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑔𝑟𝑎𝑝ℎ 𝑓 ′ 𝑥 𝑘𝑛𝑜𝑤𝑖𝑛𝑔 𝑔𝑟𝑎𝑝ℎ 𝑓(𝑥)
We can estimate the value of the derivative at any value of x by
drawing the tangent at the point (x,f(x)) and estimating its slope.
For instance, for x = 5, we draw the tangent at P in the figure
and estimate its slope to be about 3/2, so 𝑓′(5) ≈ 1.5.
This allows us to plot the point P’(5, 1.5) on the graph of f’ directly beneath P.
Repeating this procedure at several points,
we get the graph shown in this figure.
Tangents at x = A, B, and C are horizontal.
So, the derivative is 0 there and the graph
of f’ crosses the x-axis at those points.
Between A and B, the tangents have positive
slope. So, f’(x) is positive there.
Between B and C, and the tangents have
negative slope. So, f’(x) is negative there.
example: 𝑓 𝑥 = 𝑥 3 − 𝑥. Find 𝑓 ′ 𝑥 . Sketch both graphs.
  x  h 3   x  h     x 3  x 

f ( x  h)  f ( x )
 
f '( x)  lim
 lim 
h 0
h 0
h
h
x3  3x 2 h  3xh 2  h3  x  h  x3  x
 lim
h 0
h
 3x 2  1
Notice that 𝑓’(𝑥) = 0 when f has
horizontal tangents and 𝑓’(𝑥) is
positive when the tangents have
positive slope.
THE DERIVATIVE AS A FUNCTION
Example 2 b
• We use a graphing device to graph f
• and f’ in the figure.
– Notice that f’(x) = 0 when f has horizontal tangents
and f’(x) is positive when the tangents have positive
slope.
– So, these graphs serve as a check on our work in
part (a).
Two theorems:
1. Differentiability Implies Continuity
If f is differentiable at x = c, then f is continuous at x = c.
Since a function must be continuous to have a derivative, if it has a
derivative then it is continuous.
lim f ' ( x)  lim f ' ( x)  lim f ( x)  lim f ( x)  f (c)
x c 
x c
x c
x c
The converse:
"If a function is continuous at c, then it is differentiable at c," - is not true.
This happens in cases where the function "curves sharply."
𝑣𝑖𝑠𝑢𝑎𝑙𝑙𝑦: 𝑙𝑒𝑓𝑡 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 ≠ 𝑟𝑖𝑔ℎ𝑡 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒
Differentiability implies continuity,
continuity doesn’t imply differentiability.
Intermediate Value Theorem for Continuous Functions
If 𝑓(𝑥) is continuous on 𝑎, 𝑏 and 𝑘 is any
number between 𝑓(𝑎) and 𝑓(𝑏),
then there is at least one number 𝑐
such that 𝑓 𝑐 = 𝑘.
𝑓 𝑐1 = 𝑘
𝑓 𝑐2 = 𝑘
𝑓 𝑐3 = 𝑘
example:
Prove that function 𝑓 𝑥 = 𝑥2 cos(2𝑥) + 1 has a root/zero between 2 and 2.5.
𝑓 2 ≈ −1.614574
𝑓(2.5) ≈ 2.7728887
𝑓 is continuous on [2, 2.5], 𝑓(2) < 0 and 𝑓(2.5) > 0, so 𝑓 must have a zero
between 2 and 2.5.
Intermediate Value Theorem for Derivatives
Let 𝑓 be differentiable on [𝑎, 𝑏] and suppose that k is a
number between 𝑓′(𝑎) and 𝑓′(𝑏) .
Then there exists a point 𝑐 ∈ (𝑎, 𝑏) such that 𝑓′(𝑐) = 𝑘 .
You can find it in this form too:
If 𝑎 and 𝑏 are any two points in an interval on
which 𝑓 is differentiable, then 𝑓′ takes on every
value between 𝑓′(𝑎) and 𝑓′(𝑏).
Between 𝑎 and 𝑏, 𝑓′ must take on every
value between ½ and 3.
𝑓′ is continuous function on 𝑎, 𝑏
If it were always necessary to compute derivatives directly from the definition,
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
ℎ→0
𝑓 ′ 𝑥 = lim
calculus BC would be even worse nightmare, there would be no Ipad and the
life as we know it ……. Computations would be tedious, and the evaluation of
some limits would require ingenuity.
Fortunately, several rules have been developed for finding derivatives
without having to use the definition directly.
These formulas greatly simplify the task of differentiation.
First let us see what are we avoiding:
Let’s start with the simplest of all functions —
the constant function f(x) = c.
𝑓 𝑥 =𝑐
𝑓 ′ 𝑥 = lim
ℎ→0
𝑓 𝑥 + ℎ − 𝑓(𝑥)
𝑐−𝑐
= lim
= lim 0
ℎ→0
ℎ→0
ℎ
ℎ
𝑓′ 𝑥 = 0
The graph of this function is the horizontal
line y = c, which has slope 0.
In Leibniz notation
𝑑
𝑐 =0
𝑑𝑥
The derivative of a constant is zero
examples: y  3
y  0
𝑓 𝑥 = 𝑥𝑛,
𝑛 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
x  h  x

d 2
x  lim
 lim
h 0
h 0
dx
h
2
2
2
2
2
x

2
xh

h

x


h
1
1
x  h   x3

d 3
x  lim
h 0
dx
h
3
1
1
3
2
2
3
x

3
x
h

3
xh

h

x


1
2
3
 lim
h 0
d 4
x  lim
h 0
dx
h
1
 3x 2
2
3
3
We observe a pattern:
2
2
h
2
3
4
5
1
1
3
6
1
4
10 10
1
5
1
(Pascal’s Triangle)
4
x

4
x
h

6
x
h

4
xh

h

x


4
 2x
3
4
 4x 3
2x 3x 2 4x 3 5x 4 6x 5 …

We observe a pattern:
2x 3x 2 4x 3 5x 4 6x 5 …
𝑑 𝑛
𝑥 = 𝑛 𝑥 𝑛−1
𝑑𝑥
examples:
power rule
f  x  x
y  x8
4
f   x   4x
3
y  8 x 7

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑟𝑢𝑙𝑒: 𝑔 𝑥 = 𝑐𝑓(𝑥)
𝑔 𝑥+ℎ −𝑔(𝑥)
ℎ
ℎ→0
𝑔′ 𝑥 = lim
𝑐𝑓 𝑥+ℎ −𝑐𝑓(𝑥)
ℎ
ℎ→0
= lim
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
ℎ→0
= 𝑐 lim
𝑓 ′ 𝑥 = 𝑐 𝑔′(𝑥)
𝑑
𝑑
𝑐 𝑓(𝑥) = 𝑐
𝑓(𝑥)
𝑑𝑥
𝑑𝑥
examples:
d
7 x 5  7  5 x 4  35 x 4
dx
limit has no effect on a constant coefficient;
the constant could be factored to the outside.

𝑠𝑢𝑚 𝑎𝑛𝑑 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑟𝑢𝑙𝑒
𝑓 𝑥 +𝑔 𝑥
′
𝑓 𝑥 + ℎ + 𝑔 𝑥 + ℎ − 𝑓 𝑥 − 𝑔(𝑥)
= lim
ℎ→0
ℎ
𝑓 𝑥 + ℎ − 𝑓(𝑥 𝑔 𝑥 + ℎ − 𝑔(𝑥)
= lim
+
ℎ→0
ℎ
ℎ
𝑓 𝑥 + ℎ − 𝑓(𝑥)
𝑔 𝑥 + ℎ − 𝑔(𝑥)
= lim
+ lim
ℎ→0
ℎ→0
ℎ
ℎ
d
d
d
f
(
x
)

g
(
x
)

f
(
x
)

g ( x)


dx
dx
dx
limit of the sum = sum of the limits
provided they are both differentiable
each term is treated separately
examples:
y  x 4  12 x
y  x4  2x2  2
y  4 x3  12
dy
 4 x3  4 x
dx
example:
Example:
Find the horizontal tangents of:
y  x4  2x2  2
Horizontal tangents occur when slope = zero.
y  2, y  1, y  1
(The function is even, so we only get
two horizontal tangents.)
𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑟𝑢𝑙𝑒
d
d
d
 f ( x) g ( x)   f ( x)  g ( x)   g ( x)  f ( x) 
dx
dx
dx
The derivative of a product of two functions is
the first function times the derivative of the second function plus
the second function times the derivative of the first function,
provided they are both differentiable.
Example:
Find derivative if
Later, though, we will meet functions, such as
y = x2 sinx, for which the product rule is the only possible method.
Q𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑅𝑢𝑙𝑒
Q𝑢𝑜𝑡𝑖𝑒𝑛𝑡 𝑅𝑢𝑙𝑒
d  f ( x) 



dx  g ( x) 
g ( x)
d
d
f
(
x
)

f
(
x
)


 g ( x)
dx
dx
 g ( x) 2 
The derivative of a quotient 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,
provided they are both differentiable.
The theorems of this section show that:
Any polynomial is differentiable on −∞, +∞ .
Any rational function is differentiable on its domain.
Furthermore, the Quotient Rule and the other differentiation
formulas enable us to compute the derivative of any rational
function—as the next example illustrates.
Example:
Don’t use the Quotient Rule every time you see a quotient.
Sometimes, it’s easier to rewrite a quotient first
to put it in a form that is simpler for the purpose
of differentiation.
For instance:
It is possible to differentiate the function
using the Quotient Rule.
However, it is much easier to perform the division first and write the
function as
before differentiating.

𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑃𝑜𝑤𝑒𝑟 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠
The Quotient Rule can be used to extend the Power Rule
to the case where the exponent is a negative integer.
If n is a positive integer, then
d n
( x )   nx  n 1
dx
Example:
𝑃𝑜𝑤𝑒𝑟 𝑅𝑢𝑙𝑒
So far, we know that the Power Rule holds if the exponent n is
a positive or negative integer.
If n = 0, then x0 = 1, which we know has a derivative of 0.
Thus, the Power Rule holds for any integer n.
What if the exponent is a fraction?
In fact, it can be shown by using Chain Rule (obviously proof later)
that it also holds for any real number n.
If n is any real number, then
d n
n 1
( x )  nx
dx
Example:
Let
Then
y
1
3
x2
dy d 2 / 3
 (x )
dx dx
  23 x  (2 / 3) 1
  23 x 5/ 3
Example:
Find equations of the tangent line and normal line to the curve
𝑥
y=
1 + 𝑥2
at the point (1, ½).
slope of the tangent line at (1, ½) :
tangent line at (1, ½):
1
1
3
𝑦−1=− 𝑥−1 →𝑦 =− 𝑥+
4
4
4
normal line at (1, ½):
𝑦−
1
7
= 4 𝑥 − 1 → 𝑦 = 4𝑥 −
2
2
In your mind:
𝑥
′
=
1 ′
𝑥2
1 −1
1
= 𝑥 2=
2
2 𝑥
Example:
At what points on the hyperbola xy = 12 is the tangent line parallel to the line
3x + y = 0?
Since xy = 12 can be written as y = 12/x, we have:
𝑑𝑦
= 12 𝑥 −1
𝑑𝑥
′
=−
12
𝑥2
Let the x-coordinate of one of the points in question be 𝑎.
12
Slope of the tangent line at that point is − 2 , and that has to be equal to
𝑎
the slope of line 3x + y = 0
the required points are: (2, 6) and (-2, -6)
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑜𝑛 𝐹𝑜𝑟𝑚𝑢𝑙𝑎𝑠
Here’s a summary of the differentiation formulas we have learned so far.
d
c  0
dx
d n
n 1
x

nx


dx
 cf  '  cf '
 f  g  '  f ' g '
 fg  '  fg ' gf '
'
 f  g  '  f ' g '
 f  gf ' fg '
 
2
g
g
 
Higher Order Derivatives:
y 
dy
dx
is the first derivative of y with respect to x.
dy d dy d 2 y
y 

 2
dx dx dx dx
dy
y 
dx
y
 4
is the second derivative.
(y double prime)
is the third derivative.
d

y is the fourth derivative.
dx
We will learn
later what these
higher order
derivatives are
used for.
p