Transcript 2.6 The Derivative - TCC: Tidewater Community College

2.6 The Derivative
By
Dr. Julia Arnold
using Tan’s 5th edition Applied Calculus for the
managerial , life, and social sciences text
What is the derivative of something?
The derivative of a function f(x) is, mathematically
speaking, the slope of the line tangent to f(x) at any
point x.
It is also called “the instantaneous rate of change”
of a function.
It can be equated with many real world applications
such as;
Velocity which is speed such as miles per hour
In business, the derivative is called marginal, such as
the marginal Cost function, etc.
We call a line which intersects a graph in two points a
secant line. It is easy to find the slope of that line.
But it takes limits in order to find the slope of a
tangent line which touches the graph at only one point.
The green line is a secant line because it crosses the blue graph
more than once. In particular we focus on the two points
illustrated. For animation you need to be connected to the
internet.
A tangent to a graph.
To find the slope of the tangent line to a graph f(x)
we use the following formula:
h lim0
f  x  h   f ( x)
h
First let’s see how this formula equates with the
slope of a tangent line.
Notation: The derivative of f(x) is denoted by the following forms:
f  x 
dy
dx
or
f  x  h   f ( x)
h lim0
h
(x+h,f(x+h))
f ( x  h)  f ( x ) f ( x  h)  f ( x )

( x  h)  x
h
(x,f(x))
x
The slope of the
secant line would
be
x+h
By decreasing h a
little each time we
get closer and closer
to the slope of the
tangent line. By
using limits and
letting h approach 0
we get the actual
slope of the tangent
line.
The difference quotient
f x  h   f ( x)
h
measures the average rate of change of y with
respect to x over the interval [x,x+h]
In a problem pay attention to that word average.
If it is there then you do not use limits.
Differentiability and Continuity
If a function is differentiable at x = a, then it is
continuous at x = a.
However, continuous functions may contain points at
which the function is not differentiable. These are
points that create sharp points graphically, or vertical
tangent lines.
In this graph there is an abrupt
change at (a,f(a))
The slope at (a,f(a)) is
undefined.
Example 1:
Let f ( x)  x 2  6x
Find the derivative f’ of f.
Find the point on the graph of f where the tangent line to the
curve is horizontal.
Find an equation of the tangent line at the point (2, 16).
f ( x)  x 2  6 x
Find the derivative f’ of f.
f ( x  h)  f ( x )

h
( x  h ) 2  6( x  h )  x 2  6 x
h lim 0
h
x 2  2 xh  h 2  6 x  6h  x 2  6 x
h lim 0
h
x 2  2 xh  h 2  6 x  6h  x 2  6 x
h lim 0
h
2 xh  h 2  6h
h lim 0
h
h ( 2 x  h  6)
h lim 0
h
h ( 2 x  h  6)
h lim 0
h
h lim 0 2 x  h  6  2 x  6
h lim 0


Substitution of numerator.
Multiplying out
Combining like terms
Factor out h
Cancel
Let h = 0. Done.
Find the point on the graph of f where the tangent line to the curve is
horizontal.
The slope of the tangent line anywhere on the curve is given by the
formula we just found. f’(x)=2x+6
We know that the slope of a horizontal line is 0, thus
Let f’(x) = 0 and find the x that causes that.
0=2x + 6
-6 = 2x
-3=x
So the x value where the tangent line would be horizontal is -3. For
the y value substitute into the original formula
f(x) =(-3)2 +6(-3)=9-18=-9. The point is (-3,-9).
Find an equation of the tangent line at the point (2, 16).
The slope of the tangent line at this point would be found by
using the formula we found in the first part “2x+6” which would
make the slope to be 2(2)+6=10
Using the slope 10 and the point (2,16) and the point slope
formula we get
Y – 16= 10(x-2)
Y – 16 = 10x – 20
Y =10x-4
Find the average rate of change for f(x) = x2 - 3x + 4 from
x = 2 to x = 3, from x = 2 to x = 2.5, from x = 2 to x = 2.1.
When finished find the instantaneous rate of change at
x = 2.
Solution: Use the derivative formula without doing the limit.
Thus simplify just the f(x  h)  f(x) part of the formula.
h
(x  h)  3(x  h)  4  (x  3x  4)
h
x2  2xh  h2  3x  3h  4  x2  3x  4
h
x2  2xh  h2  3x  3h  4  x2  3x  4
h
2xh  h2  3h
h
h(2x  h  3)
h
2x  h  3
2
2x  3  h
2
So f(x) = x2 - 3x + 4 average rate of change formula simplifies to
2x - 3 + h.
First Interval
x = 2 to x = 3 h is the difference between the two values or 3-2=1
Second Interval
x = 2 to x = 2.5 h = .5
Third Interval
x = 2 to x = 2.1 h = .1
Since the second number in the interval is actually x + h in the
formula, that makes x = 2 in this problem.
First Interval x = 2 and h = 1. Substitute into 2x - 3 + h and the
average rate of change is 2(2) - 3 + 1 = 2
Second Interval x = 2 and h = .5 Substitute 2(2)-3 +.5 = 1.5
Third Interval x = 2 and h = .1 Substitute 2(2)-3 + .1 = 1.1
As you can see as we get closer to x = 2 the slopes are decreasing
toward 1. For the instantaneous rate of change, we use the limit
as h approaches 0 to obtain 2x - 3 at x = 2 which is 1.
Similar problems to the ones illustrated here will be
on Test 1.