Transcript The Derivative and the Tangent Line Problem

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.
P
Note: Although tangent lines do
not intersect a circle, they may
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.
Why would we want a tangent
f
line???
P
Remember, the closer you zoom in on point
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.
Definition of Derivative:
This is a
major part of
calculus and
we will
differentiate
until the cows
come home!
Also,
(provided the limit exists,)
f ( x  x)  f ( x)
f '( x)  lim
x 0
x
Really a
fancy slope
formula…
change in y
divided by
the change in
x.
= slope of the line tangent to the graph of f at
(x, f(x)).
= instantaneous rate of change of f(x) with
respect to x.
Definition of the Derivative of a
Function
Notations For Derivative
Let y  f(x)
df ( x ) dy

 Df ( x)  y '
f '( x) 
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
d
f  x  does not mean
times f  x  !
dx
dx
(except when it is convenient to treat it that way.)

4
3
2
y  f  x
1
0
3
The derivative
is the slope of
the original
function.
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
0
-1
-2
1
2
3
4
5
6
7
8
9
y  f   x

6
5
y  x 3
2
4
3
2
1
-3
-2
-1 0
-1
1
x
2
3
y  lim
 x  h
2
h0
-2


3 x 3
2
h
-3
6
5
4
3
2
1
-3 -2 -1 0
-1
-2
-3
-4
-5
-6
x  2 xh  h  x
y  lim
h 0
h
0
y  lim 2 x  h
2
1 2 3
x
2
2
h 0
y  2 x

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  0 x
x  0
x
'
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  0 x
x  0
x
'
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  0 x
x  0
x
'
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  0 x
x  0
x
'
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  0 x
x  0
x
'
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x2  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)  x2  3 , find f ’(x) and the
equation of the tangent lines at:
a) x = 1
b) x = -2
a) x = 1:
2
2
(
x


x
)

3

(
x
 3)
'
f ( x)  lim
x 0
x
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x2  3 , find f ’(x) and the
equation of the tangent lines at:
a) x = 1
b) x = -2
a) x = 1:
2
2
(
x


x
)

3

(
x
 3)
'
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)  x2  3 , find f ’(x) and the
equation of the tangent lines at:
a) x = 1
b) x = -2
a) x = 1:
2
2
(
x


x
)

3

(
x
 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 0
x
x
The Slope of the Graph of a
Non-Linear Function
Example: Given f ( x)  x2  3 , find f ’(x) and the
equation of the tangent lines at:
a) x = 1
b) x = -2
a) x = 1:
2
2
(
x


x
)

3

(
x
 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 0
x
x
 lim (2 x  x)  2 x
x 0
The Slope of the Graph of a Non-Linear
Function
2
Example: Given f ( x)  x  3, find f ’(x) and the
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  0
x
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)  x2  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 0
x
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

'
x


x
x
f ( x)  lim
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
x  x  x
1
1

x( x  x)
'
x


x
x
f ( x)  lim
 lim 
x  0
x  0
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
x  x  x
1
1

x( x  x)
'
x


x
x
f ( x)  lim
 lim 
x  0
x  0
x
x
x
1
 lim
 2
x  0 xx ( x  x )
x
Example-Continued
x  x  x
1
1

x( x  x)
'
x


x
x
f ( x)  lim
 lim 
x  0
x  0
x
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)
f ( x)  lim
x 0
x
3
'
3
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  x3 )  3x  3x  2 x3  3x
 lim 
x 0
x
'
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.
3  3( x x)  (2 x3  3x)
2(
x

x
)
lim
x0
x
3  2 x2x  4 x2x  4 xx2  2 xx2  2x3)  3x  3x  2 x3  3x
(2
x
 lim
x0
x
2  6 xx  2x2  3)

x
(6
x
 lim
x0
x
f '( x) 
Derivative
Example: Find the derivative of f(x) = 2x3 – 3x.
3  3( x x)  (2 x3  3x)
2(
x

x
)
lim
x0
x
3  2 x2x  4 x2x  4 xx2  2 xx2  2x3)  3x  3x  2 x3  3x
(2
x
 lim
x0
x
2  6 xx  2x2  3)

x
(6
x
 lim
x0
x
 6 x2  3
f '( x) 
Derivative
Example: Find f '( x) for f ( x)  x
Derivative
Example: Find f '( x) for f ( x)  x
f ( x)  lim
'
x 0
x  x  x ( x  x  x )

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
on your calculator.
3
4
2
Select 5: Tangent(
Type the x value, which in
this case is 1, and then hit

(I changed my window)
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
at x = 2 does not exist.
-5
5
-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.
Line: 3x – y – 4 = 0
f(x) = x3 + 2
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
Taking my word for it, the derivative of the function is
This is 3 when x is
3x
2
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
– no pole
– no sharp corner
A derivative is a limit !