Aim: Basic Differentiation Course: Calculus Do Now: Aim: What are some of the basic rules of...

Aim: Basic Differentiation Course: Calculus

Do Now:

Aim: What are some of the basic rules of differentiation?

On what interval(s) is the derivative of f(x) increasing?2


Definition of the Derivative

The derivative of f at x is

h

xfhxfxf

h

)()(lim)('

0

provided this limit exists.

The derivative f’(x) is a formula for the slope of the tangent line to the graph of f at the point (x,f(x)).

The function found by evaluating the limit of the difference quotient is called the derivative of f at x. It is denoted by f ’(x), which is read “f prime of x”.


Alternate Notation for Derivative

( )d

f xdx

xD y

dy

dx

2( )d x

dx

. . . read as “the derivative of y with respect to x.”

0

( ) ( )'( ) lim

h

f x h f xf x

h

0

( ) ( )' lim

h

f x h f xy

h


The derivative of a constant function is 0. That is, if c is a real number, then

The Constant Rule

dc

dx0

y = 7

function derivative

f(x) = 0 s(t) = -3

dydx

0

f’(x) = 0 s’(t) = 0


If n is a rational number, then the function f(x) = xn is differentiable and

For f to be differentiable at x = 0, n must be a number such that xn-1 is defined on an interval containing 0.

The Power Rule

nnd

nxxdx

1

f(x) = x3

function derivative

y = 1/x2

g x x3( ) d

g x x xdx x

1 23 3

23

1 1'( ) [ ]

3 3

dy dx x

dx dx x2 3

3

2[ ] 2

Note: dx

dx1

4

2

-2

-4

q x = x

f x x2'( ) 3


Model Problems

Find the slope of the graph f(x) = x4 using the power rule when x = 0.

f(x) = x4

f’(x) = 4x3

f’(0) = 4(0)3 = 0

nnd

nxxdx

1 Power Rule

evaluate d/dx for 0


Model Problems

Find the equation of the tangent line to the graph f(x) = x2 when x = -2.

f(x) = x2

f’(x) = 2x

f’(0) = 2(-2) = -4 = m - the slope of tangent line at (2,-4)

nnd

nxxdx

1 Power Rule

y – y1 = m(x – x1) Point Slope form

y – 4 = -4[x – (-2)] Substitute for y1,, m, and x1

y = -4x – 4 Simplify


The Constant Multiple Rule

If f is a differentiable function and c is a real number, then cf is also differentiable and d

cf xcf xdx

'( )( )

y = 2/x

function

dy d dx x

dx dx dx

xx

1 1

22

22

22 1

derivative

yx3 2

1

2

532

3

53

1 212 32

1

3

dy dxx

dx dx

x


The Sum & Difference Rules

The derivative of the sum (or difference) of two differentiable functions is differentiable and is the sum (or difference) of their derivatives.

df x g xf x g x

dx'( ) '( )( ) ( )

df x g xf x g x

dx'( ) '( )( ) ( )

Sum Rule

Difference Rule

f(x) = -x4/2 + 3x3 – 2x

function

derivative

f’(x) = -2x3 + 9x2 – 2

g(x) = x3 – 4x + 5 g’(x) = 3x2 – 4


Do Now:

Aim: What are some of the basic rules of differentiation?

Find the equation of the tangent line for f(x) = 4x3 - 7x2, at x = 3.


Sine and Cosine

The derivatives of the Sine and Cosine Functions:

dxx

dxcossin

dxx

dxsincos

y = 2 sin x

function derivative

y = x + cos x

xy

sin2

x

y x1 cos

' cos2 2

y’ = 2 cos x

y x' 1 sin


Recall: Finding Slope at Point

3

2.5

2

1.5

1

0.5

1

x, f(x)

3

2.5

2

1.5

1

0.5

1

(x + h, f(x + h))

h

f(x + h) – f(x)

3

2.5

2

1.5

1

0.5

1

x, f(x)

h

f(x + h) – f(x)

(x + h, f(x + h))

3

2.5

2

1.5

1

0.5

1

x, f(x)

h

f(x + h) – f(x)

(x + h, f(x + h))

3

2.5

2

1.5

1

0.5

1

sec0

tan lim mmslopeh

h

xfhxfm

h

)()(lim

0

difference quotient

the first derivative of the function


x

ymslope

sec

h

xfhxfmslope

)()(sec

Average Rate of Change

3

2.5

2

1.5

1

0.5

1

3

2.5

2

1.5

1

0.5

1

(x + h, f(x + h))

h

f(x + h) – f(x)

x, f(x)


Average Rate of Change

The average rate of change of a function f(x) on an interval [a, b] is found by computing the slope of the line joining the end points of the function on that interval.

average rate of change

of ( ) on interval [ , ]

( ) ( )

f x a b

f b f a

b a

Find avg. rate of change f(x) = x2 for [-1, 2]

(2) ( 1)1

2 ( 1)

f f

3

2.5

2

1.5

1

0.5

1

3

2.5

2

1.5

1

0.5

1

(b, f(b))

b - a

f(b) – f(a)

a, f(a)


Rates of Change

Following is graph of f with various line segments connecting points on the graph.

The average rate of change of f over the interval [.5, 2] is the slope of which line segment?

CD


Rates of Change

D = rt

If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function s = -16t2 + 100 where s is measured in feet and t is measured in seconds. Find the average velocity over the time interval [1, 1.5].

Average velocity = st

Change in distanceChange in time

Position Function s t gt v t s20 0

1( )

2

g = -32 ft/sec2

-40 ft/sec


3

2.5

2

1.5

1

0.5

1

x, f(x)

3

2.5

2

1.5

1

0.5

1

(t + h, s(t + h))

h

s(t + h) – s(t)

t, s(t)

Instantaneous Velocity – Time & Distance

3

2.5

2

1.5

1

0.5

1

t, s(t)

h

s(t + h) – s(t)

(t + h, s(t + h))

3

2.5

2

1.5

1

0.5

1

t, s(t)

h

s(t + h) – s(t)

(t + h, s(t + h))

3

2.5

2

1.5

1

0.5

1

Velocity Function

t

s st t tv tt0

( ) lim

032t v

the first derivative of the Position Function

s t gt v t s20 0

1( )

2

is g = -32 ft/sec2time

dis

tan

ce


Model Problems

At time t = 0, a diver jumps from a diving board that is 32 feet above the water. The position of the diver is given by

where s is measured in feet and t is measured in seconds.

a. When does the diver hit the water?

b. What is the diver’s velocity at impact?

s t t t2( ) 16 16 32

a. diver hits water when s = 0. Substitute and solve for t.

t t20 16 16 32 t t0 16( 1)( 2)

t or1 2

2 sec.


Model Problems

At time t = 0, a diver jumps from a diving board that is 32 feet above the water. The position of the diver is given by

where s is measured in feet and t is measured in seconds.

a. When does the diver hit the water?

b. What is the diver’s velocity at impact?

s t t t2( ) 16 16 32

2 sec.

b. s’(t) = -32t + 16 nnd

nxxdx

1 Power Rule

s’(t) = -32(2) + 16 = -48 Substitute 2 from a.

-48 feet per second

Aim: Basic Differentiation Course: Calculus Do Now: Aim: What are some of the basic rules of...

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