• Find the slope of the tangent line to a curve at a point.• Use the limit definition to find the derivative of a function.• Understand the relationship between differentiability and continuity.

2.1 The Derivative and the Tangent Line Problem

The need for calculus:

1. To find the changing slope for a curve2. To find velocity and acceleration for moving

objects3. To find maximum and minimum points4. To find area of an irregular shape

This section will focus on finding slope of a tangent line.

4

2

-2

5

The tangent line to a circle is the line perpendicular to the radius through the point.

A tangent is ???Touching but not crossing the curve?Touching curve at only one point?

In 1.1 we approximated the slope of the tangent line through a point on a curve by approximating the slope of the secant line through the point and a nearby point. (Review powerpoint 1.1 if needed).

msec=f(c+ x) - f(c)

(c+ x) - c

m=y2 - y1

x2 - x1

The denominator is Δx, the small change in x, and Δy = f(c+Δx) – f(c) is the small change in y.

msec=f(c+ x) - f(c)

x

As you let Δx get infinitely small, through the limiting process, you get the slope.

The slope of the tangent line to the graph of f at the point (c, f(c)) is called the slope of the graph of f at x=c

Ex 1 p. 98 The slope of the graph of a linear function

Find the slope of the graph of f(x) = 3x – 5 at the point (2, 1)

10

8

6

4

2

5

y=3

x=1

f x = 3x-5

0 0

(2 ) (2) [3(2 ) 5] [3 2 5]lim limx x

f x f x

x x

0

[6 3 5] [6 5]limx

x

x

0

3limx

x

x

3

But we should already have known this, right?

Ex 2 p. 98 Tangent lines to the graph of nonlinear functions

f(x) = x2 + 1

To avoid doing the same kind of thing twice, let (c, f(c)) be some arbitrary point on graph.

Find the slopes for

at (0, 1) and (-1, 2)

2 2

0 0

( ) ( ) [( ) 1] [( ) 1]lim limx x

f c x f c c x c

x x

2 2 2

0

[ 2 ( ) ( ) 1] [ 1]limx

c c x x c

x

2 2 2

0

2 ( ) ( ) 1 1limx

c c x x c

x

2

0

2 ( ) ( )limx

c x x

x

0lim 2 2x

c x c

So for (0, 1), c=0 and m=2٠0; for (-1, 2), c= -1 and m=2٠-1

This definition doesn’t cover possibility of vertical tangent lines. For vertical tangent lines, if f is continuous at c, and

0

( ) ( )limx

f c x f c

x

or

0

( ) ( )limx

f c x f c

x

Then you know you have a vertical tangent line.

Now we are ready for a big breakthrough . . . (drum roll please)

Be sure to notice that the derivative also results in a function, f ′(x), which is read “f prime of x”

Other notations for f ′(x) are:dy

dx, y',

d

dx[f(x)], Dx [y]

The process of finding a derivative is called differentiation.

dy/dx is read as the derivative of y with respect to x, or dy dx

Ex 3 p. 100 Finding derivative by the limit processFind the derivative of f(x) = x3 - 5x

0

( ) ( )'( ) lim

x

f x x f xf x

x

3 3

0

[( ) 5( )] [ 5 ]limx

x x x x x x

x

2 33 2 3

0

3 3 5 5 5limx

x x x x x x x x x x

x

2 32

0

3 3 5limx

x x x x x x

x

22

0lim (3 3 5)x

x x x x

2'( ) 3 5f x x This is a function which can output slopes for any x chosen on the graph of f(x); In other words, a function machine!

Ex 4 p 100 Using derivative to find the slope at a point.Find f '(x) for f(x) = x

Then find the slope of f at points (1,1), (4,2). Describe the behavior of f at (0,0)

0

( ) ( )'( ) lim

x

f x x f xf x

x

0

( )limx

x x x

x

0

( ) ( )lim

( )x

x x x x x x

x x x x

0

( )lim

[ ( ) ]x

x x x

x x x x

0

1lim

( )x x x x

1'( )

2f x

x 0x

1 1'(1) , '(4) , '(0)

2 4f f f undefined

Ex 5 p. 101 Finding the derivative of a function

Find the derivative with respect to t for the function y = 3/t. Find dy/dt

4

3

2

1

5

g x = 3

x

0 0

3 3( ) ( ) ( )

lim limt t

dy f t t f t t t tdt t t

Multiply by LCD of all denominators, t(t+Δt) as “one”

0

( )3 31( )

lim( )

1t

t t tt t t

t t tt

0

3lim

( )( )t

t

t t t t

2

3dydt t

0

3 3( )lim

( )( )t

t t t

t t t t

Differentiability and Continuity

If a function is not continuous at x=c it is also not differentiable at x=c.

( ) ( )'( ) lim

x c

f x f cf c

x c

Alternate form: derivative at x=c is

( ) ( )limx c

f x f c

x c

( ) ( )limx c

f x f c

x c

and

provided the one-sided limits

exist and are equal.

It is true that differentiability implies continuity, but the converse is not necessarily true. If a function is continuous at a point, it is not always differentiable at that point. Examples 6, 7 will demonstrate that.

Ex 6 p. 102 Graph with sharp turn

f(x) = | x-3 |

8

6

4

2

5

m=-1m=1

f x = x-3

this is continuous at x=3but not differentiable at x=3

3 3

3 3 3( ) (3)lim lim 1

3 3x x

xf x f

x x

3 3

3 3 3( ) (3)lim lim 1

3 3x x

xf x f

x x

These one-sided limits are not equal, so f(x) is not differentiable.

Ex 7 p. 102 Graph with vertical tangent line

2

1

-1

-2

-5 5

f x = x1

3

This is continuous at x=0, but

13

0 0

( ) (0) 0lim lim

0x x

f x f x

x x

20 3

1limx x

The tangent line is vertical and the function is not differentiable at x=0

Assign. 2.1 p.103/ 1-45 EOO, 57, 63, 71,73, 75, 81-85 odd