Math 127H: Lecture 5 Definition of the Derivative

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Math 127H: Lecture 5

Definition of the Derivative

Example 1 .

We consider the graph of a function, say y = x 2 − x + 1 and a point on the graph, say P = ( x = 2 , y = 3) , and ask: What is the equation of the tangent to the curve at P ?

We do not solve this problem directly. We find better and better approximations to the tangent line and take the limit of these lines. This gives us the tangent line.

Example 2 .

Let s denote position and t denote time. Assume we are given s = 100 − 32 t

2 and ask, what is the velocity at time t = 4?

If we compute the average velocity between time t = 4 + h and time t = 4 we obtain different results depending on h.

We notice that as h gets smaller and smaller these approximations approach a fixed number. We call this number the instantaneous velocity.

Definition 1.

Given a function f :

R at x = p is the

R

, and a value lim h → 0 f ( p + h ) − f ( p ) h x = p for x , the derivative of f provided the limit exists.

Notation: Let y = f ( x ) be a function. The derivative of f at x = p is denoted dy dx

( p ) = df dx

( p ) = f

0

( p ) .

Example

3

.

We give another physical interpretation of the derivative. First we draw a function as arrows from the domain to the codomain ( or range). We can, with a little poetic license, think of the arrows as light rays and the function as a lens. The derivative is local magnification. See the figure below.

We compute the derivative directly from the definition. Later we develop algorithms that allow us to handle many functions in a simpler fashion.

Example 4 .

Let f ( x ) = 3 x

2 − 5 x + 3 .

Then f ( x + h ) − f ( x )

= h

(3( x + h )

2 − 5( x + h ) + 3) − (3 x

=

6 xh + 3 h 2 − 5 h h h

2 − 5 x + 3)

We compute df dx

= f

0

( x ) = lim h → 0

6 xh + 3 h

2 − 5 h h

= lim h → 0

(6 x + 3 h − 5) why? = 6 x − 5 .

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Example 5 .

We compute the derivative of g ( t ) =

3 t − 2

2 t +7

.

g ( t + h ) − g ( t ) h

= (

3( t + h ) − 2

2( t + h ) + 7

3 t − 2

) /h

2 t + 7

=

(3( t + h ) − 2)(2 t + 7) − (3 t − 2)(2( t + h ) + 7) h (2 t + 7)(2(( t + h ) + 7)

25 h

= h (2 t + 7)(2( t + h ) + 7)

When we take the limit of this expression as h → 0 we cancel the h in the denominator with the h in the numerator. We obtain g

0

( t ) =

25

(2 t + 7) 2

.

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Example 6 .

We compute, by hand, the derivative of f ( x ) =

√ x.

√ x + h −

√ x h

=

(

√ x + h − h (

√ x )(

√ x + h + x + h +

√ x )

√ x )

= h (

√ x + h − x x + h +

√ x )

=

1

√ x + h +

√ x if h = 0

Taking the limit as h → 0 we obtain df dx

=

2

1

√ x

.

Application 1.

Often physical or biological phenomena can be described locally. By locally I mean over short distances , short periods of time, or . Assume that a population

P ( t ) grows at a constant rate R . This means that the number of new individuals in a time interval h is equal to a percent of the population at the beginning of the time interval.

This says that the population change is

P ( t + h ) − P ( t )

= P ( t ) R.

h

To use this we pick h and assume that P ( t ) is constant over the time interval from t to t + h.

But P ( t ) may change over that time interval. Hence we replace the above statement with dP

= RP.

dt

Of course this is also problematical.

Application 2.

We return to the chemical reaction

A + B −→ AB.

Let x ( t ) denote the number of units of AB in the container at time t.

We assume that x (0) = 0, the amount of A at time t = 0 is a , and the amount of B at time t = 0 is b.

We have that the rate of the reaction is

R ( t ) ∝ amount of A and the amount of B or dx

= k ( x − a )( x − b ) .

dt

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Application 3.

Let F denote force, v denote velocity, a denote acceleration, and m denote mass. The derivative of instantaneous velocity is instantaneous acceleration, or dx

= v.

dt

In We drop the word instantaneous henceforth.

We have F = ma.

This is well and good if the force is constant. For example if it depends on position, then, without using derivatives, the statement does not apply. An example is working in a gravitational field that varies from point to point. Assume that the force of gravity is − 1 /x 2 .

Then we have the meaningful statement:

F = ma, or

− 1 /x

2

= m dv dt

.

In the above applications we have examples of differential equations.

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