Top Banner
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
19

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.

Dec 26, 2015

Download

Documents

Ruth Mitchell
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: 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.

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

Page 2: 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.

Aim: Basic Differentiation Course: Calculus

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”.

Page 3: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 4: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 5: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 6: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 7: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 8: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 9: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 10: 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.

Aim: Basic Differentiation Course: Calculus

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.

Page 11: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 12: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 13: 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.

Aim: Basic Differentiation Course: Calculus

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)

Page 14: 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.

Aim: Basic Differentiation Course: Calculus

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)

Page 15: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 16: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 17: 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.

Aim: Basic Differentiation Course: Calculus

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

Page 18: 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.

Aim: Basic Differentiation Course: Calculus

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.

Page 19: 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.

Aim: Basic Differentiation Course: Calculus

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