THE GRAPH OF THEDERIVATIVE

Section 3.4 / 3.5

January 21, 2013

TANGENT LINES

Suppose we have the graph of a function f (x).

X Axis

Y A

xis

y = f (x)

a

(a, f(a))

Tangent line at x = a

b

(b, f(b))

NOT a tangent line at x = b

The tangent line of the graph of f at a is the line that:

(1) touches the graph of f at the point (a, f (a)) and intersects itat no other point near a,

(2) gives the “direction of the curve”.

TANGENT LINES

Intuitively:

Think of driving a car on the graph. The line that your head lights

and tail lights make is the tangent line to the graph at that point.

X Axis

Y A

xis

y = f (x)

THE EQUATION OF THE TANGENT LINE

First, we need to determine is the slope of the tangent line.

Let f be a function whose graph is given below, and supposewe want to determine the slope of the tangent line of the graphof f at x = a.

a

(a, f(a))

a+h

(a+h, f(a+h))

Consider the secant lines from (a, f (a)) to (a + h, f (a + h))

THE EQUATION OF THE TANGENT LINE

We see that the slope of the secant lines approaches the slope ofthe tangent line as h ! 0. So...

Slope of the Tangent Line

Slope of the tangent lineof f at x = a

= limh!0

f (a + h)� f (a)

h

= f

0(a)

Now that we know the slope of the tangent line, we can use thepoint-slope form of a line to determine its equation:

Equation of the Tangent LineThe equation of the line tangent to the graph of f at x = a is

y = f

0(a)(x � a) + f (a)

Provided f

0(a) exists.

EXAMPLES

Determine the equation of the tangent line to the graph of thefollowing functions at the points given.

(a) f (x) = x

2 + 2x

at x = 3(b) g(t) = 5/t

at t = 2(c) h(x) = 4

px

at x = 9

(a) First we determine f

0(3).

f

0(3) = limh!0

(3 + h)2 + 2(3 + h)� (32 + 2(3))h

= limh!0

9 + 6h + h

2 + 6 + 2h � 9 � 6h

= limh!0

(8 + h) = 8

Since f

0(3) = 8 and f (3) = 15, the equation of the tangent line is:

y = 8(x � 3) + 15 = 8x � 9

EXAMPLES

Determine the equation of the tangent line to the graph of thefollowing functions at the points given.

(a) f (x) = x

2 + 2x

at x = 3(b) g(t) = 5/t

at t = 2(c) h(x) = 4

px

at x = 9

(b) First we determine g

0(2).

g

0(2) = limh!0

52+h

� 52

h

= limh!0

10�5(2+h)2(2+h)

h

= limh!0

�5h

2h(2 + h)= lim

h!0

�52(2 + h)

= � 54

Since g

0(2) = �5/4 and g(2) = 5/2, the equation of the tangentline is:

y = �54(x � 2) +

52

= � 54

x + 5

EXAMPLES

Determine the equation of the tangent line to the graph of thefollowing functions at the points given.

(a) f (x) = x

2 + 2x

at x = 3(b) g(t) = 5/t

at t = 2(c) h(x) = 4

px

at x = 9

(c) First we determine h

0(9).

h

0(9) = limh!0

4p

9 + h � 4p

9h

= limh!0

4p

9 + h � 12h

= limh!0

(4p

9 + h � 12)h

· (4p

9 + h + 12)(4p

9 + h + 12)

= limh!0

16(9 + h)� 144h(4

p9 + h + 12)

= limh!0

164p

9 + h + 12=

1624

=23

So, y =23(x � 9) + 12 =

23

x + 6.

EXISTENCE OF THE DERIVATIVE

As mentioned last class, it is easier to determine if thederivative of a function at a point exists by looking at its graph.

Existence of the DerivativeThe derivative of a function f at a point x = a exists if all of thefollowing are satisfied:

(1) f is continuous at x = a.(2) f is smooth at x = a (the graph has no sharp corners)(3) f does NOT have a vertical tangent line at x = a.

If ANY ONE of these conditions is not satisfied, then thederivative of f at x = a does not exist.

EXAMPLES

X Axis

Y A

xis

a b c d e f

(x = a) Derivative does NOT exist (not continuous)(x = b) Derivative does NOT exist (not continuous)(x = c) Derivative does NOT exist (vertical tangent)(x = d) Derivative does NOT exist (not continuous)(x = e) Derivative exists!(x = f ) Derivative does NOT exist (not smooth)

EXAMPLES

Graph each of the following functions. Determine whether thederivative of the function at x = 0 exists, if it does not, explainwhich condition fails.

(a) f (x) = x

2/3 (b) g(t) = t

1/3 (c) h(x) = x

�2/3

Does NOT exist

Vertical tangent

at x = 0

Does NOT exist

Not smooth

at x = 0

Does NOT exist

Not continuous

at x = 0

ESTIMATING THE DERIVATIVE FROM THE GRAPH

In the real world you often find yourself with a graph of afunction, but not the equation describing it.

Below is the graph of the distance traveled from Omaha (miles)travelled in the 1.5 hours of a road trip to Minnesota.

0.5 1 1.5 2 2.5

10

20

30

40

50

60

70

80

90

X Axis

Y A

xis

Estimate the speed of the car 0.5 hours into the trip. (⇡ 40 mph)

ESTIMATING THE DERIVATIVE FROM THE GRAPH

In the real world you often find yourself with a graph of afunction, but not the equation describing it.

Below is the graph of the distance traveled from Omaha (miles)travelled in the 1.5 hours of a road trip to Minnesota.

0.5 1 1.5 2 2.5

10

20

30

40

50

60

70

80

90

X Axis

Y A

xis

Estimate the speed of the car 1 hour into the trip. (0 mph)

EXAMPLES

Below is the graph of the function g(x).

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12 13

-11

-8

-6

-4

-2

2

4

6

8

X Axis

Y A

xis

Use this graph to approximate the following:

(a) g

0(1) = � 2 (b) g

0(4) = 0.5 (c) g

0(7) = 4.5

SKETCHING THE DERIVATIVE

Let’s take what we just did one step further and sketch thegraph of a derivative from the graph of the original function.

Below is the graph of a function f (x). Sketch the graph of f

0(x).

-6 -5 -4 -3 -2 -1 1 2 3 4 5 6

-4

-3

-2

-1

1

2

3

4

5

X Axis

Y A

xis

SKETCHING THE DERIVATIVE

Below is the graph of a function f (x). Sketch the graph of f

0(x).

SKETCHING THE DERIVATIVE

Below is the graph of a function g(t). Sketch the graph of g

0(t).

SKETCHING THE DERIVATIVE

Below is the graph of a function h(x). Sketch the graph of h

0(x).