Average rate of change Find the rate of change if it takes 3 hours to drive 210 miles. What is your...

Post on 22-Dec-2015

219 views 0 download

Transcript of Average rate of change Find the rate of change if it takes 3 hours to drive 210 miles. What is your...

Average rate of Average rate of changechange

Find the rate of change if it takes 3 hours to drive 210 miles.

What is your average speed or velocity?

( ) (3 0

3

)

0

f f

If it takes 3 hours to drive 210 miles If it takes 3 hours to drive 210 miles

then we averagethen we average

A.A. 1 mile per minute1 mile per minute

B.B. 2 miles per minute2 miles per minute

C.C. 70 miles per hour70 miles per hour

D.D. 55 miles per hour55 miles per hour

If it takes 3 hours to drive 210 miles If it takes 3 hours to drive 210 miles

then we averagethen we average

A.A. 1 mile per minute1 mile per minute

B.B. 2 miles per minute2 miles per minute

C.C. 70 miles per hour70 miles per hour

D.D. 55 miles per hour55 miles per hour

Instantaneous slopeInstantaneous slope

What if h went to What if h went to zero?zero?

0'( ) l

( (im

) )h

f x h x

hf x

f

DerivativeDerivative

if the limit exists as one real if the limit exists as one real number. number.

0'( ) l

( (im

) )h

f x h x

hf x

f

DefinitionDefinitionIf f : D -> K is a function then the derivative of f If f : D -> K is a function then the derivative of f

is a new function, is a new function, f ' : D' -> K' as defined above if the limit f ' : D' -> K' as defined above if the limit

exists. exists. Here the limit exists every where except at x = 1Here the limit exists every where except at x = 1

0'( ) l

( (im

) )h

f x h x

hf x

f

Guess at Guess at

0

1lim

) 1( ( )h

hf

h

f

Guess at Guess at

0

1lim

) 1( ( )h

hf

h

f

EvaluateEvaluate

0

1lim

) 1( ( )h

hf

h

f

EvaluateEvaluate

1.51.5

0.50.50

1lim

) 1( ( )h

hf

h

f

EvaluateEvaluate

0

1lim

) 1( ( )h

hf

h

f

EvaluateEvaluate

-1 = -1 = 0

1lim

) 1( ( )h

hf

h

f

ThusThus

==

0'( ) li

1 1m

( ) ( )h

f fhf

hx

ThusThus

d.n.e.d.n.e.

0'( ) li

1 1m

( ) ( )h

f fhf

hx

Guess at Guess at

f’(0.2) – slope of f when x = f’(0.2) – slope of f when x = 0.20.2

0'( ) l

( (im

) )h

f x h x

hf x

f

Guess at f ’(3)Guess at f ’(3)

Guess at f ’(3)Guess at f ’(3)

-1.0-1.0

0.490.49

Guess at f ’(-2)Guess at f ’(-2)

Guess at f ’(-2)Guess at f ’(-2)

-3.0-3.0

1.991.99

Note that the rule is Note that the rule is f '(x) is the slope at the point ( x, f(x) ), f '(x) is the slope at the point ( x, f(x) ), D' is a subset of D, butD' is a subset of D, butK’ has nothing to do with KK’ has nothing to do with K

0'( ) l

( (im

) )h

f x h x

hf x

f

K is the set of distances from homeK is the set of distances from homeK' is the set of speeds K' is the set of speeds K is the set of temperaturesK is the set of temperaturesK' is the set of how fast they rise K' is the set of how fast they rise K is the set of today's profits , K is the set of today's profits , K' tells you how fast they changeK' tells you how fast they changeK is the set of your averages K is the set of your averages K' tells you how fast it is changing. K' tells you how fast it is changing.

0'( ) l

( (im

) )h

f x h x

hf x

f

Theorem If f(x) = c where c Theorem If f(x) = c where c is a real number, then f ' (x) is a real number, then f ' (x) = 0.= 0.

Proof : Lim [f(x+h)-f(x)]/h = Proof : Lim [f(x+h)-f(x)]/h =

Lim (c - c)/h = 0.Lim (c - c)/h = 0.

Examples Examples

If f(x) = 34.25 , then f ’ (x) = 0If f(x) = 34.25 , then f ’ (x) = 0

If f(x) = If f(x) = , then f ’ (x) = 0, then f ’ (x) = 0

If f(x) = 1.3 , find f’(x)If f(x) = 1.3 , find f’(x)

0.00.0

0.10.1

Theorem Theorem If f(x) = x, then f ' (x) = 1. If f(x) = x, then f ' (x) = 1.

Proof : Lim [f(x+h)-f(x)]/h = Proof : Lim [f(x+h)-f(x)]/h =

Lim (x + h - x)/h = Lim h/h = 1Lim (x + h - x)/h = Lim h/h = 1

What is the derivative of x What is the derivative of x grandson?grandson?

One grandpa, one.One grandpa, one.

Theorem If c is a constant,Theorem If c is a constant,(c g) ' (x) = c g ' (x) (c g) ' (x) = c g ' (x)

Proof : Lim [c g(x+h)-c g(x)]/h =Proof : Lim [c g(x+h)-c g(x)]/h =

c Lim [g(x+h) - g(x)]/h = c g ' (x) c Lim [g(x+h) - g(x)]/h = c g ' (x)

Theorem If c is a constant,Theorem If c is a constant,(cf) ' (x) = cf ' (x) (cf) ' (x) = cf ' (x)

( 3 x)’ = 3 (x)’ = 3 or( 3 x)’ = 3 (x)’ = 3 or

If f(x) = 3 x then If f(x) = 3 x then

f ’(x) = 3 times the derivative of xf ’(x) = 3 times the derivative of x

And the derivative of x is . . And the derivative of x is . .

One grandpa, one !!One grandpa, one !!

If f(x) = -2 x then f ’(x) If f(x) = -2 x then f ’(x) = =

-2.0-2.0

0.10.1

TheoremsTheorems

1. (f + g) ' (x) = f ' (x) + g ' (x), and 1. (f + g) ' (x) = f ' (x) + g ' (x), and

2. (f - g) ' (x) = f ' (x) - g ' (x) 2. (f - g) ' (x) = f ' (x) - g ' (x)

1. (f + g) ' (x) = f ' (x) + g ' 1. (f + g) ' (x) = f ' (x) + g ' (x) (x) 2. (f - g) ' (x) = f ' (x) - g ' 2. (f - g) ' (x) = f ' (x) - g ' (x) (x)

If f(x) = 3If f(x) = 322 x + 7, find f ’ x + 7, find f ’ (x)(x)

f ’ (x) = 9 + 0 = 9f ’ (x) = 9 + 0 = 9

If f(x) = x - 7, find f ’ (x)If f(x) = x - 7, find f ’ (x)

f ’ (x) = - 0 = f ’ (x) = - 0 =

55 5

If f(x) = -2 x + 7, find f ’ If f(x) = -2 x + 7, find f ’ (x)(x)

-2.0-2.0

0.10.1

If f(x) = thenIf f(x) = then f’(x) = f’(x) =

Proof : f’(x) = Lim [f(x+h)-f(x)]/h = Proof : f’(x) = Lim [f(x+h)-f(x)]/h =

x1

2 x

g(x) = 1/x, find g’(x)g(x) = 1/x, find g’(x)

g(x+h) = 1/(x+h)g(x+h) = 1/(x+h) g(x) = 1/xg(x) = 1/x

g’(x) = g’(x) =

1 1 ( )1

( )1

x x hx h x

x x hh

1 1x h xh

( )

( )

x x h

hx x h

1

2

1

x

If f(x) = xIf f(x) = xnn then f ' (x) = n x then f ' (x) = n x (n-(n-

1)1)

If f(x) = xIf f(x) = x44 then f ' (x) = 4 xthen f ' (x) = 4 x33

If If 2

3( )g x

x 23x

2 2 3'( ) (3 ) ' 3( ) ' 3( 2 )g x x x x 3

3

66x

x

If f(x) = xIf f(x) = xnn then f ' (x) = n x then f ' (x) = n xn-1 n-1

If f(x) = xIf f(x) = x44 + 3 x+ 3 x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4

f ' (x) = 4 xf ' (x) = 4 x3 3 + . . . .+ . . . .

f ' (x) = 4xf ' (x) = 4x33 + 9 x+ 9 x22 - 4 x – 3 + 0 - 4 x – 3 + 0

f(1) = 1 + 3 – 2 – 3 + 4 = 3f(1) = 1 + 3 – 2 – 3 + 4 = 3

f ’ (1) = 4 + 9 – 4 – 3 = 6f ’ (1) = 4 + 9 – 4 – 3 = 6

3y

If f(x) = xIf f(x) = xnn then f ' (x) = n x then f ' (x) = n x (n-(n-

1)1)If f(x) = If f(x) = xx44 then f ' (x) = 4then f ' (x) = 4 x x33

If f(x) = If f(x) = 44 then f ' (x) = 0then f ' (x) = 0 If If ( ) 3g x x

1

23x1 1 1

2 2 21

'( ) (3 ) ' 3( ) ' 3( )2

g x x x x

1

23 3

2 2x

x

If f(x) = then f ‘(x) =If f(x) = then f ‘(x) =x

1 1

2 21

'( ) ( ) '2

f x x x

1

2 x

Find the equation of the line Find the equation of the line tangent to g when x = 1. tangent to g when x = 1.

If g(x) = xIf g(x) = x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4

g ' (x) = 3 xg ' (x) = 3 x22 - 4 x – 3 + 0 - 4 x – 3 + 0

g (1) =g (1) =

g ' (1) =g ' (1) =

If g(x) = xIf g(x) = x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4find g (1)find g (1)

0.00.0

0.10.1

If g(x) = xIf g(x) = x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4find g (1)find g (1)

If g(x) = xIf g(x) = x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4find g’ (1)find g’ (1)

If g(x) = xIf g(x) = x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4find g’ (1)find g’ (1)

-4.0-4.0

0.10.1

Find the equation of the Find the equation of the line tangent to f when x line tangent to f when x = 1. = 1.

g(1) = 0g(1) = 0

g ' (1) = – 4g ' (1) = – 4

14

0

x

y

4(0 1)xy

( 1)4y x

Find the equation of the line Find the equation of the line tangent to f when x = 1. tangent to f when x = 1.

If f(x) = xIf f(x) = x44 + 3 x+ 3 x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4

f ' (x) = 4xf ' (x) = 4x33 + 9 x+ 9 x22 - 4 x – 3 + 0 - 4 x – 3 + 0

f (1) = 1 + 3 – 2 – 3 + 4 = 3f (1) = 1 + 3 – 2 – 3 + 4 = 3

f ' (1) = 4 + 9 – 4 – 3 = 6 f ' (1) = 4 + 9 – 4 – 3 = 6

Find the equation of the Find the equation of the line tangent to f when x line tangent to f when x = 1. = 1.

f(1) = 1 + 3 – 2 – 3 + 4 = 3f(1) = 1 + 3 – 2 – 3 + 4 = 3

f ' (1) = 4 + 9 – 4 – 3 = 6 f ' (1) = 4 + 9 – 4 – 3 = 6

61

3Y

X

Write the equation of the Write the equation of the tangent line to f when x = 0. tangent line to f when x = 0.

If f(x) = xIf f(x) = x44 + 3 x+ 3 x33 - 2 x - 2 x22 - 3 x + 4 - 3 x + 4

f ' (x) = 4xf ' (x) = 4x33 + 9 x+ 9 x22 - 4 x – 3 + 0 - 4 x – 3 + 0

f (0) = write downf (0) = write down

f '(0) = for last questionf '(0) = for last question

Write the equation of the Write the equation of the line tangent to f(x) when x line tangent to f(x) when x = 0.= 0.A.A. y - 4 = -3xy - 4 = -3x

B.B. y - 4 = 3xy - 4 = 3x

C.C. y - 3 = -4xy - 3 = -4x

D.D. y - 4 = -3x + 2y - 4 = -3x + 2

Write the equation of the Write the equation of the line tangent to f(x) when x line tangent to f(x) when x = 0.= 0.A.A. y - 4 = -3xy - 4 = -3x

B.B. y - 4 = 3xy - 4 = 3x

C.C. y - 3 = -4xy - 3 = -4x

D.D. y - 4 = -3x + 2y - 4 = -3x + 2

http://www.youtube.com/watch?v=P9dpTTpjymE Derive Derive

http://www.9news.com/video/player.aspx?aid=52138&bw= Kids= Kids

http://math.georgiasouthern.edu/~bmclean/java/p6.html Secant Lines Secant Lines

Find the derivative of Find the derivative of each of the following. each of the following. 3.13.1

52

3

2 3( ) 3 2

4 2 72( )

2

f x x xx x

x xg x

x

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t)=16.3tN(t)=16.3t0.87660.8766.. How many sold in 2001?How many sold in 2001? How fast was sales increasing in How fast was sales increasing in

2001?2001? How many sold in 2005?How many sold in 2005? How fast was sales increasing in How fast was sales increasing in

2005?2005?

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t)=16.3tN(t)=16.3t0.87660.8766.. How many sold in 2001?How many sold in 2001? N(1)= 16.3 million camera soldN(1)= 16.3 million camera sold

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t) =16.3tN(t) =16.3t0.87660.8766

How fast was sales increasing in How fast was sales increasing in 2001?2001?

N’(t) =N’(t) =

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t) =16.3tN(t) =16.3t0.87660.8766

How fast was sales increasing in How fast was sales increasing in 2001?2001?

N’(t) = 0.8766*16.3tN’(t) = 0.8766*16.3t-0.1234-0.1234

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t) =16.3tN(t) =16.3t0.87660.8766

How fast was sales increasing in How fast was sales increasing in 2001?2001?

N’(t) = 0.8766*16.3tN’(t) = 0.8766*16.3t-0.1234-0.1234

N’(1) = 14.2886 million per yearN’(1) = 14.2886 million per year

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t)=16.3tN(t)=16.3t0.87660.8766.. How many sold in 2005?How many sold in 2005? N(5)= 66.8197 million cameras N(5)= 66.8197 million cameras

soldsold

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t) =16.3tN(t) =16.3t0.87660.8766

How fast was sales increasing in How fast was sales increasing in 2005?2005?

N’(t) =N’(t) =

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t) =16.3tN(t) =16.3t0.87660.8766

How fast was sales increasing in How fast was sales increasing in 2005?2005?

N’(t) = 0.8766*16.3tN’(t) = 0.8766*16.3t-0.1234-0.1234

53. Millions of cameras53. Millions of cameras t=1 means 2001 t=1 means 2001 N(t) =16.3tN(t) =16.3t0.87660.8766

How fast was sales increasing in How fast was sales increasing in 2005?2005?

N’(t) = 0.8766*16.3tN’(t) = 0.8766*16.3t-0.1234-0.1234

N’(5) = .8766*16.3/5N’(5) = .8766*16.3/50.12340.1234

11.7148 million per year11.7148 million per year

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

Find the velocity for any t.Find the velocity for any t. Find the velocity when brakes Find the velocity when brakes

applied.applied. When did it stop?When did it stop?

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

Find the velocity for any t.Find the velocity for any t. x’(t) = 120 - 30 t x’(t) = 120 - 30 t

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

x’(t) = 120 - 30 t x’(t) = 120 - 30 t Find the velocity when brakes Find the velocity when brakes

applied.applied. x’(0) = 120 ft/secx’(0) = 120 ft/sec

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

x’(t) = 120 - 30 t x’(t) = 120 - 30 t When did it stop?When did it stop?

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

x’(t) = 120 - 30 t x’(t) = 120 - 30 t When did it stop?When did it stop? x’(t) = 120 - 30 t = 0 x’(t) = 120 - 30 t = 0

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

x’(t) = 120 - 30 t x’(t) = 120 - 30 t When did it stop?When did it stop? x’(t) = 120 - 30 t = 0 x’(t) = 120 - 30 t = 0 120 - 30 t = 0120 - 30 t = 0 120 = 30 t120 = 30 t 4 = t4 = t

Dist trvl by X-2 racing Dist trvl by X-2 racing carcart seconds after braking. t seconds after braking. 59.59. x(t) = 120 t – 15 tx(t) = 120 t – 15 t22..

x’(t) = 120 - 30 t x’(t) = 120 - 30 t When did it stop?When did it stop? t = 4 sec.t = 4 sec. x(4) = 480 – 15*16 = 240 feet x(4) = 480 – 15*16 = 240 feet