Finding f '(x) from the Definition of the Derivative (page 643)

The four steps used to find the derivative f ' (x) for a function y = f (x)

Are summarized here.

1. Find f( x + h ).

Steps: Example: f (x) = 6x 2 – 28x + 15

f( ) = 6( ) 2 – 28( ) + 15

f( x + h ) = 6( x + h ) 2 – 28( x + h ) + 15

f( x + h ) = 6( x 2 + 2xh + h 2 ) – 28( x + h ) + 15

f( x + h ) = 6x 2 + 12xh + h 2 – 28x – 28h + 15

2. Find and simplify

f ( x + h ) – f ( x ).

f ( x + h ) – f ( x ) = ( 6x 2 + 12xh + h 2 – 28x – 28h + 15 ) – ( 6x 2 – 28x + 15 )

f ( x + h ) – f ( x ) = 6x 2 + 12xh + h 2 – 28x – 28h + 15 – 6x 2 + 28x – 15

f ( x + h ) – f ( x ) = 12xh + h 2 – 28h

3. Divide by h to get

h

xfhxf )()( h

hhhx

h

xfhxf 2812

)()( 2

h

hxh )2812(

2812 hx

4. Let h → 0;

h

xfhxf

x

)()(

0lim

if this limit exists.

)2812( )()(

00

limlim

hx

hh

xfhxf

h

28)0(12 x

2812 x

Note: These are the steps you must use and follow to work problem 20 on

your test. I am a stickler for following instructions so get use to it.

An easier way to have found the first derivative would have been to simply use the rules from the next chapter. You may use this short cut to find the derivative in problem 33 on the test.

f ( x ) = 6x 2 – 28x + 15

f ' ( x ) = 6( 2 ) x 2 – 1 – 28 x 1 – 1

f ' ( x ) = 12 x – 28 x 0

f ' ( x ) = 12 x – 28

From algebra: x 0 = 1

Multiply the exponent times the coefficient and subtract one from the existing exponent. The derivative of a constant is zero.

The prime mark on f is important and must be written. It is one of the notations used to signify the first derivative.

Remember that the first derivative yields rate of change.

So if we evaluate f ' ( x ) at x = 2 for the example from the previous slide

we would obtain the following information: f ' ( 2 ) = 12 ( 2 ) – 28 = – 4.

The negative four indicates the function was decreasing at a rate of four units when x was 2.

If we were interested in determining what was happening when x was 3,

f ' ( 3 ) would yield 12 ( 3 ) – 28 which is 36 – 28 making 8 the rate of change.

The positive eight indicates the function was increasing at a rate of eight units when the variable was three.

You may use this short cut to work problem 33 but I want to see the various steps in determining if the function is decreasing or increasing at the point of interest.