4. Learning Objectives;
Students will be able to learn about;
What is Icrement.
What is meant by change in x i.e.(∆x).
What is meant by chane in y i.e .(∆y) .
What is average rate of change.
What is incremental ratio.
What is limit of function .
What is derivative of function.
5. What is Increment
The change in the values of ‘y’corresponding to
small changes in the values of ‘x’ where ‘y’ &’x’
are related to each other by the equation;
let y=x^3.when x=2 we have y=8.
if x changes from 2 to 2.1.Then increment
The corresponding value of y will be
Now if x changes from 2 to 0.5.Then increment in
x=0.5-2=-1.5 Corresponding value of y at x=0.5 will
be=0.125 Increment in y in this caseis
Hence it is concluded now a small change in the value of
x as it increases or decreases from one value of
X=x1 to another value of X=x2.
in its change is denoted by ∆x. (read as’Deltax’). And is called increment
Here ∆x= x2-x1 and we write x2=x1 + ∆x .
Note. ∆x is not product of ∆ & x but merely a symbol for a small
change in x.
7. The limit of a function
A quantity L is the limit of a function f (x) as
x approaches a if, as the input values of x
approach a (but do not equal a), th
corresponding output values of f (x) get
closer to L. Note that the value of the limit
is not affected by the output value of f (x) at
a. Both a and L must be real numbers. We
write it as
lim f (x) = L
x → a
8. Incremental Ratio
Consider the function y=f(x).whose domain isreal
As x changes to x+∆x,there will be a
change in y of amount,say ∆y.
Therefore,y will change from y to y+ ∆y.
∆y.= f(x+∆x)-y (by subtracting y
or. ∆y.= f(x+∆x)-f(x). Eq(1).
9. Icremental ratio contd.
Now dividing both side equation (1). By ∆x we
The above ratio of increment in y
the increment in x is called the incremental
ratio or the difference quotient.
Now we learn the pattern of change in the ratio
corresponding to change in x.
Let us take initially x=3.
we take ∆x=4,3,2,1,0.1,0.01,0.001,……..etc.
The following table gives the corresponding of
and that of .
View table on next slide in detail.
From above table we observe that ∆x decreases
and approaches 0,∆y also dcreases and
approaches 0.But the icremental ratio does not
approach 0.instead it approaches a definite value
Hence we can conclude that
lim = 6
x → a
13. Derivative of f(x).
The limit to the ratio as ∆x tends to 0.is called
the derivative or differential coeffcient of y w.r.to
Definition; Let y=f(x) be a continuous function of ‘
∆x → 0 i.e. , if it
is called the differential co-efficient of f(x) w.r.t.’x’and
denoted by dy/dx, d/dx(y), Dxy, d/dx(f(x)).
14. How to…
Given a function f, find the derivative by
applying the definition of the derivative.
1. Calculate f (a + h).
2. Calculate f (a).
3. Substitute and simplify f (a + h) − f (a)
4. Evaluate the limit if it exists:
f′(a) = lim f (a + h) − f (a)
h →0 h
Finding the Derivative of a Polynomial
Find the derivative of the function
f (x) = x2 − 3x + 5 at x = a.
Solution We have
f′(a) = lim f (a + h) − f (a) [Definition of a
h →0 h
Substitute f (a + h) = (a + h)2 − 3(a + h) + 5
f (a) = a2 − 3a + 5 in the formula
23. Try It #3
Find the derivative of the function
f (x) = 10x + 11 ,at x=a
5x + 4
24. Finding Derivatives of Functions
To find derivatives of functions with roots,
we use the methods we have learned to find
limits of functions with roots,including
multiplying by a conjugate.
25. Example 4;
Finding the Derivative of a Function with a
Find the derivative of the function f (x) = 4√ x
at x = 36.
Solution; We have
27. Try It #4
Find the derivative of the function f (x) = 9√ x
at x = 9.
by following the same above procedure.
28. Finding the Average Rate of Change of
The functions describing the examples above
involve a change over time. Change divided by time
is one example of a rate. The rates of change in the
previous examples are each different. In other
words, some changed faster than others.
If we were to graph the functions, we could compare
the rates by determining the slopes of the graphs.
29. Concept of a tangent to a curve
A tangent line to a curve is a line that
intersects the curve at only a single point but
does not cross it there. (The tangent
line may intersect the curve at another point
away from the point of interest.) If we zoom in
on a curve at that point,
the curve appears linear, and the slope of the
curve at that point is close to the slope of the
tangent line at that point.
Figure represents the function f (x) = x3 − 4x.
We can see the slope at various points along
• slope at x = −2 is 8
• slope at x = −1 is –1
• slope at x = 2 is 8
31. Explanation with Graph
Let’s imagine a point on the curve of function f
at x = a as shown in Figure . The coordinates of
the point are (a, f (a)). Connect this point with a
second point on the curve a little to the right of x
= a, with an x-value increased by some small
real number h.
The coordinates of this second point are
((a + h, f (a + h)) for some positive-value h.
We can calculate the slope of the line
connecting the two points (a, f (a)) and (a + h, f
(a + h)), called a secant line, by
applying the slope formula, slope = change
change in x
We use the notation msec to represent the
slope of the secant line connecting two points.
35. Example 1;
Finding the Average Rate of Change
Find the average rate of change connecting
the points (2, −6) and (−1, 5).
Solution; We know the average rate of change
connecting two points may be given by
If one point is (2, − 6), or (2, f (2)), then f (2) = −6.
The value h is the displacement from 2 to −1, which
equals −1 − 2 = −3.
For the other point, f (a + h) is the y-coordinate at a +
h, which is 2 + (−3) or −1, so f (a + h) = f (−1) = 5.
37. Try It #1
Find the average rate of change connecting
the points (−5, 1.5) and ( − 2.5, 9).?.
38. Understanding the Instantaneous
Rate of Change
Now that we can find the average rate of
change, suppose we make h in Figure 2
smaller and smaller. Then a + h will approach a
as h gets smaller, getting closer and closer to
0. Likewise, the second point (a + h, f (a + h))
the first point, (a, f (a)). As a consequence, the
connecting line between the two points, called
the secant line, will get closer and closer to
being a tangent to the function at x = a, and the
slope of the secant line will get closer and
closer to the slope of the tangent at x = a. See
40. Explanation of IROC
Because we are looking for the slope of the tangent at x = a, we can
think of the measure of the slope of the curve of
a function f at a given point as the rate of change at a particular
instant. We call this slope the instantaneous rate of change, or
the derivative of the function at x = a. Both can be found by finding
the limit of the slope of a line connecting the point at x = a with a
second point infinitesimally close along the curve. For a function f
both the instantaneous rate of change of the function and the
derivative of the function at x = a are written as f′(a), and we
candefine them as a two-sided limit that has the same value
whether approached from the left or the right.
The expression by which the limit is found is known
as the difference quotient.
41. Finding Instantaneous Rates of Change
Many applications of the derivative involve
determining the rate of change at a given instant of a
function with the
independent variable time—which is why the term
instantaneous is used. Consider the height of a ball
with an initial velocity of 64 feet per second, given
s(t) = −16t2 + 64t + 6, where t is measured in
seconds and s(t)is measured in feet. We know the
path is that of a parabola. The derivative will tell us
how the height is changing atany given point in
time. The height of the ball is shown in Figure as a
function of time. In physics, we call this the
43. Example ;
Finding the Instantaneous Rate of
Using the function above, s(t) = −16t2 + 64t
+ 6, what is the instantaneous velocity of
the ball at 1 second and 3
seconds into its flight?
Solution ;The velocity at t = 1 and t = 3 is
the instantaneous rate of change of
distance per time, or velocity. Notice that
the initial height is 6 feet. To find the
instantaneous velocity, we find the
s′(t) = − 32t + 64 Evaluate the limit by letting h
For any value of t, s′(t) tells us the velocity at
that value of t.
Evaluate t = 1 and t = 3.
s′(1) = −32(1) + 64 = 32
s′(3) = −32(3) + 64 = −32
The velocity of the ball after 1 second is 32
feet per second, as it is on the way up.
The velocity of the ball after 3 seconds is −32
feet per second, as it is on the way down.
46. Try It #5
The position of the ball is given by
s(t) = −16t2 + 64t + 6. What is its velocity
after 2 seconds into flight?
47. Using Graphs to Find
Instantaneous Rates of Change
We can estimate an instantaneous rate of
change at x = a by observing the slope of the
curve of the function f (x) at x = a. We do this
by drawing a line tangent to the function at x
= a and finding its slope.
48. How To…
Given a graph of a function f (x), find the
instantaneous rate of change of the function at
x = a.
1. Locate x = a on the graph of the function f
2. Draw a tangent line, a line that goes through
x = a at a and at no other point in that section
of the curve. Extendthe line far enough to
calculate its slope as
change in y
Estimating the Derivative at a Point on the
Graph of a Function
From the graph of the function y = f (x)
presented in Figure , estimate each of the
f (0); f (2); f′(0); f′(2)
50. Solution ;
To find the functional value, f (a), find the y-
coordinate at x = a.
To find the derivative at x = a, f′(a), draw a
tangent line at x = a, and estimate the slope
of that tangent line. See Figure.
51. a. f (0) is the y-coordinate at x = 0. The point has
coordinates (0, 1), thus f (0) = 1.
b. f (2) is the y-coordinate at x = 2. The point has
coordinates (2, 1), thus f (2) = 1.
c. f′(0) is found by estimating the slope of the tangent
line to the curve at x = 0. The tangent line to the curve
x = 0 appears horizontal. Horizontal lines have a slope
of 0, thus f′(0) = 0.
d. f′(2) is found by estimating the slope of the tangent
line to the curve at x = 2. Observe the path of the
tangent line to the curve at x = 2. As the x value moves
one unit to the right,
52. Try It #6
Using the graph of the function f (x) = x3 − 3x
shown in Figure , estimate: f (1), f′(1), f (0), and