# The Derivative and the Tangent Line Problem - PowerPoint - PowerPoint

Document Sample

```					The Derivative and the
Tangent Line Problem
Section 2.1
After this lesson, you should be
able to:
• find the slope of the tangent line to a curve
at a point
• use the limit definition of a derivative to
find the derivative of a function
• understand the relationship between
differentiability and continuity
Tangent Line
A line is tangent to a curve at a point P if the line is
perpendicular to the radial line at point P.

Note: Although tangent lines do
not intersect a circle, they may
P
cross through point P on a curve,
depending on the curve.
The Tangent Line Problem
Find a tangent line to the graph of f at P.
f       Why would we want a tangent
line???
Remember, the closer you zoom in on point
P             P, the more the graph of the function and the
tangent line at P resemble each other. Since
finding the slope of a line is easier than a
curve, we like to use the slope of the tangent
line to describe the slope of a curve at a point
since they are the same at a particular point.
A tangent line at P shares the same point and
slope as point P. To write an equation of any
line, you just need a point and a slope. Since
you already have the point P, you only need to
find the slope.
Definition of
a Tangent
• Let Δx shrink
from the left

f ( x0  x)  f ( x0 )
m  lim
x 0           x
Definition of a Tangent Line with
Slope m
The Derivative of a Function
Differentiation- the limit process is used to define the slope
of a tangent line.
Really a
fancy slope
Definition of Derivative:    (provided the limit exists,)    formula…
f ( x  x)  f ( x)   change in y
This is a                     f '( x)  lim                          divided by
x 0          x            the change in
major part of    Also,                                               x.
calculus and
= slope of the line tangent to the graph of f at
we will
(x, f(x)).
differentiate
until the cows           = instantaneous rate of change of f(x) with
come home!               respect to x.
Definition of the Derivative of a
Function
Notations For Derivative

Let y  f(x)

df ( x) dy
f '( x)              Df ( x)  y '
dx      dx

If the limit exists at x, then we say that f is differentiable at x.
dx does not mean d times x !

dy does not mean d times y !


dy
does not mean dy  dx !
dx
(except when it is convenient to think of it as division.)

df
does not mean df  dx !
dx
(except when it is convenient to think of it as division.)


d
f  x  does not mean d
times f  x  !
dx                       dx
(except when it is convenient to treat it that way.)


4

3

2

1
y  f  x

0    1      2     3     4    5     6    7     8    9
3
The derivative
is the slope of   2
the original               The derivative is defined at the end points
function.         1        of a function on a closed interval.

0    1      2     3      4   5     6     7    8        9

-1
y  f  x
-2

6
5
4
3
y  x 3  2

           
2

 x  h
2
1
3 x 3     2
-3 -2 -1 0
-1
1
x
2   3
y  lim
-2                       h 0                   h
-3
6
5
x  2 xh  h  x
2                 2           2
4
3
y  lim
h 0         h
2
1
0
y  lim 2 x  h
-3 -2 -1 0    1 2 3
-1     x
-2                                     h0
-3

y  2 x
-4
-5
-6                                                                       
A function is differentiable if it has a
derivative everywhere in its domain. It
must be continuous and smooth.
Functions on closed intervals must have
one-sided derivatives defined at the end
points.

p
Theorem 2.1 Differentiability Implies
Continiuty
The Slope of the Graph of a
Line
Example: Find the slope of the graph of f ( x)  2 x  1
at the point (2, 5).
The Slope of the Graph of a
Line
Example: Find the slope of the graph of f ( x)  2 x  1
at the point (2, 5).

2( x  x)  1  (2 x  1)
f ( x)  lim
'
x 0            x
2 x  2x  1  2 x  1          2x
 lim                              lim        lim 2  2
x  0            x              x  0 x   x  0
The Slope of the Graph of a
Line
Example: Find the slope of the graph of f ( x)  2 x  1
at the point (2, 5).

2( x  x)  1  (2 x  1)
f ( x)  lim
'
x 0            x
2 x  2x  1  2 x  1          2x
 lim                              lim        lim 2  2
x  0            x              x  0 x   x  0
The Slope of the Graph of a
Line
Example: Find the slope of the graph of f ( x)  2 x  1
at the point (2, 5).

2( x  x)  1  (2 x  1)
f ( x)  lim
'
x 0            x
2 x  2x  1  2 x  1          2x
 lim                              lim        lim 2  2
x  0            x              x  0 x   x  0
The Slope of the Graph of a
Line
Example: Find the slope of the graph of f ( x)  2 x  1
at the point (2, 5).

2( x  x)  1  (2 x  1)
f ( x)  lim
'
x 0            x
2 x  2x  1  2 x  1          2x
 lim                              lim        lim 2  2
x  0            x              x  0 x   x  0
The Slope of the Graph of a
Line
Example: Find the slope of the graph of f ( x)  2 x  1
at the point (2, 5).

2( x  x)  1  (2 x  1)
f ( x)  lim
'
x 0            x
2 x  2x  1  2 x  1          2x
 lim                              lim        lim 2  2
x  0            x              x  0 x   x  0
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x 2  3 , find f ’(x) and the
equation of the tangent lines at:

a) x = 1

b) x = -2

a) x = 1:
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x 2  3 , find f ’(x) and the
equation of the tangent lines at:

a) x = 1

b) x = -2
( x  x)  3  ( x  3)
2            2

a) x = 1:   f ( x)  lim
'
x 0           x
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x 2  3 , find f ’(x) and the
equation of the tangent lines at:

a) x = 1

b) x = -2
( x  x) 2  3  ( x 2  3)
a) x = 1:         f ' ( x)  lim
x  0             x
x 2  2 x x   x 2  3  x 2  3
 lim
x  0                 x
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x 2  3 , find f ’(x) and the
equation of the tangent lines at:

a) x = 1

b) x = -2
( x  x) 2  3  ( x 2  3)
a) x = 1:   f ( x)  lim
'
x  0             x
x 2  2 xx  x 2  3  x 2  3              x (2 x  x )
 lim                                          lim
x  0                 x                     x  0      x
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x 2  3 , find f ’(x) and the
equation of the tangent lines at:

a) x = 1

b) x = -2

a) x = 1:                     ( x  x) 2  3  ( x 2  3)
f ' ( x)  lim
x 0             x
x 2  2 xx  x 2  3  x 2  3            x (2 x  x )
 lim                                          lim
x  0                x                     x 0      x
 lim (2 x  x)  2 x
x  0
The Slope of the Graph of a Non-Linear
Function
Example: Given f ( x)  x  3, find f ’(x) and the
2

equation of the tangent lines at:
a) x = 1
a) x = 1:
b) x = -2
( x  x) 2  3  ( x 2  3)
f ( x)  lim
'
x 0             x
x 2  2 xx  x 2  3  x 2  3            x(2 x  x)
 lim                                        lim
x 0                x                     x 0     x
 lim (2 x  x)  2 x
x 0

At X = 1, 2x = 2. So, the slope of the Tangent line is 2, and the equation of the tangent line is:
y  2 x  b. When, x is 1, y is 4, so use (1, 4), to find 4 = 2(1) + b --> b = 2.
so, y  2 x  2 is the equation of the tangent line at x = 1.
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x 2  3 , find f ’(x) and the
equation of the tangent line at:
b) x = -2

( x  x)2  3  ( x 2  3)
f ( x)  lim
'
x 0            x
x 2  2 xx  x 2  3  x 2  3           x(2 x  x)
 lim                                       lim
x 0                x                    x 0     x
 lim (2 x  x)  2 x
x 0

At X = -2, 2x = -4. So, the slope of the Tangent line is -4, and the equation of the tangent line is:
y  4 x  b. When, x is -2, y is , so use (4, 7), to find 7 = -2(-4) + b --> b = -1.
so, y  4 x  1 is the equation of the tangent line at x = -2.
The Slope of the Graph of a
Non-Linear Function
Example: Find f ’(x) and the equation of the tangent line at
x = 2 if          1
f ( x) 
x
The Slope of the Graph of a
Non-Linear Function
Example: Find f ’(x) and the equation of the tangent line at
x = 2 if          1
f ( x) 
x

1     1

f ( x)  lim
'            x  x x
x 0     x
The Slope of the Graph of a
Non-Linear Function
Example: Find f ’(x) and the equation of the tangent line at
x = 2 if          1
f ( x) 
x

1     1       x  x  x

f ( x)  lim
'            x  x x  lim  x( x  x)
x 0     x     x 0     x
The Slope of the Graph of a
Non-Linear Function
Example: Find f ’(x) and the equation of the tangent line at
x = 2 if          1
f ( x) 
x

1     1        x  x  x

f ( x)  lim
'              x  x x  lim  x( x  x)
x 0      x      x 0     x
x          1
 lim                    2
x 0 xx ( x  x )   x
Example-Continued
1     1        x  x  x

f ( x)  lim
'              x  x x  lim  x( x  x)
x 0      x      x 0     x
x          1
 lim                    2
x 0 xx ( x  x )   x
If x = 2, the slope is, -¼. So, y = 1/4x + b. Going back to the
original equation of y = 1/x, we see if x = 2, y = 1/2. So:

1 1                                            1
    2   b  b  1, and the equation is y  x  1
2   4                                           4
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.

2( x  x)  3( x  x)  (2 x  3x)
3                      3
f ( x)  lim
'
x 0                  x
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.

2( x  x)3  3( x  x)  (2 x 3  3x)
f ( x)  lim
'
x  0                   x
2( x3  x 2 x  2 x 2 x  2 xx 2  xx 2  x 3 )  3x  3x  2 x 3  3x
 lim 
x  0                                       x
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.

f '( x) 
lim 2( x x)3  3( x x)  (2x3  3x)
x0                  x
 lim (2x3  2x2x  4x2x  4xx2  2xx2  2x3)  3x  3x  2x3  3x
x0                                   x
 lim x(6x2  6xx  2x2  3)
x0             x
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.

f '( x)   lim 2( x x)3  3( x x)  (2x3  3x)
x0                  x
 lim    (2x3  2x2x  4x2x  4xx2  2xx2  2x3)  3x  3x  2x3  3x
x0                                     x
 lim    x(6x2  6xx  2x2  3)
x0               x
 6 x2  3
Derivative
Example: Find f '( x) for f ( x)  x
Derivative
Example: Find f '( x) for f ( x)  x

x  x  x ( x  x  x )
f ( x)  lim
'

x 0         x      ( x  x  x )
Derivative
Example: Find f '( x) for f ( x)  x

x  x  x ( x  x  x )
f ( x)  lim
'

x 0       x       ( x  x  x )
x  x  x
 lim
x  0 x ( x  x     x)
Derivative
Example: Find f '( x) for f ( x)  x

x  x  x ( x  x  x )
f ( x)  lim
'

x  0      x        ( x  x  x )
x  x  x                  x          1
 lim                         lim                  
x  0 x ( x  x     x ) x 0 x( x  x  x ) 2 x

THIS IS A HUGE
RULE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Example-Continued
Let’s work a little more with this example…
Find the slope of the graph of f at the points (1, 1) and (4, 2). What
happens at (0, 0)?

1
, at (1, 1) the slope is 1/2.
2 x
at (4, 2) the slope is 1/4. However, at (0,0)
this slope is undefined!!!!!!!!!!!!!!!!!!!!!
Example-Continued
Let’s graph tangent lines with our calculator…we’ll draw the
tangent line at x = 1.

1 Graph the function f ( x)  x          3      Select 5: Tangent(
4     Type the x value, which in
this case is 1, and then hit

(I changed my window)
2
Now, hit  DRAW

Here’s the equation of the tangent
line…notice the slope…it’s approximately
what we found
Differentiability Implies
Continuity
If f is differentiable at x, then f is continuous at x.

Some things which destroy differentiability:
1. A discontinuity (a hole or break or asymptote)
2. A sharp corner (ex. f(x)= |x| when x = 0)
3. A vertical tangent line (ex: f ( x )    3
x   when x = 0)
2.1
Differentiation Using Limits
of Difference Quotients
• Where a Function is Not
Differentiable:

• 1) A function f(x) is not differentiable at a
point x = a, if there is a “corner” at a.
2.1
Differentiation Using Limits
of Difference Quotients
•   Where a Function is Not Differentiable:

•   2) A function f (x) is not differentiable at a point
•      x = a, if there is a vertical tangent at a.
3. Find the slope of the tangent line to f  x   x  2 at x = 2.

4
This function has a sharp turn at x = 2.
2

Therefore the slope of the tangent line
-5         5
at x = 2 does not exist.
-2

-4

Functions are not differentiable at
a. Discontinuities
b. Sharp turns
c. Vertical tangents
2.1
Differentiation Using Limits
of Difference Quotients
Where a Function is Not Differentiable:

3) A function f(x) is not differentiable at a point x = a, if it is not continuous at a.

Example: g(x) is not
continuous at –2,
so g(x) is not
differentiable at x = –2.
1
4. Find any values where f  x        is not differentiable.
x 3
4
This function has a V.A. at x = 3.
2

Therefore the derivative at x = 3 does
not exist.                                 -5            5

-2

-4

Theorem:
If f is differentiable at x = c,
then it must also be continuous at x = c.
Example

Find an equation of the line that is tangent to the graph of f
and parallel to the given line.
f(x) = x3 + 2           Line: 3x – y – 4 = 0
Example
Find an equation of the line that is tangent to the graph of f
and parallel to the given line.
f(x) = x3 + 2                 Line: 3x – y – 4 = 0
Parallel to 3x  y  4  0, implies the slope of the tangent line must be the same as
the slope of the line. So, solving for y, we get
y  3x  4, so the slope = 3
2
Taking my word for it, the derivative of the function is   3x
This is 3 when x is
1
If x = 1, y = 3. If x = -1, y = 1. So there are two possible lines.
The first is 3 = 3(1)+b  y  3x
The second is 1= 3(-1) +b  y = 3x + 4
Definition of Derivative
• The derivative is the formula which
gives the slope of the tangent line at
any point x for f(x)
f ( x0  x)  f ( x0 )
f '( x)  lim
x 0           x
• Note: the limit must exist
– no hole
– no jump                   A derivative is a limit !
– no pole
– no sharp corner

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 56 posted: 3/25/2012 language: pages: 54
How are you planning on using Docstoc?