1007 : the shortcut

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1007 : The Shortcut AP CALCULUS

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1007 : The Shortcut. AP CALCULUS. Notation:. The Derivative is notated by:. Newton . L’Hopital. Leibniz. Derivative of the function. With respect to x. Notation used in Cal 3. Notation:. Find the rate of change of the Circumference of a Circle with respect to its Radius. - PowerPoint PPT Presentation

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Page 1: 1007  : The Shortcut

1007 : The Shortcut

AP CALCULUS

Page 2: 1007  : The Shortcut

Notation:

The Derivative is notated by:

( )

x

y y f x

dy d y Ddx dx

Newton L’Hopital

LeibnizDerivative of the function

With respect to x Notation

used in Cal 3

Page 3: 1007  : The Shortcut

Notation:Find the rate of change of the Circumference of a Circle with respect to its Radius.

Find the rate of change of the Area of a Square with respect to the length of a Side.

Find the rate of change of the Volume of a Cylinder with respect to its Height.

L’Hopital

Leibniz

A´(s)𝑑 𝑨𝑑 𝒔

𝑽 ′ (𝒉)V

𝐴=𝒔2

You treat r as a constant

C

𝑑𝑽𝑑𝒉=

Page 4: 1007  : The Shortcut

Algebraic Rules

REM: A). A Constant Function

(3)ddx

( )( )

f x cf x

0

Derivative is the slope of a tangent

y = 3 m = 0

0

Page 5: 1007  : The Shortcut

Algebraic Rules

B). A Power Function( )( )

nf x xf x

32

1( ) df x x y xdx x

Rewrite in exponent form!

𝒏𝒙𝒏−𝟏

𝒇 ′ (𝒙 )=𝟑 𝒙𝟐()

y

𝒚 ′ (𝒙 )=𝟏𝟐 𝒙−𝟏𝟐

y’ (x) ==-2

Page 6: 1007  : The Shortcut

( )( )

nf x xf x

𝒏𝒙𝒏−𝟏

𝒚= 𝟏𝟑√𝒙

Rewrite in exponent form!

¿ 𝒙−𝟏𝟑

𝒚 ′ (𝒙 )=−𝟏𝟑 𝒙

−𝟒𝟑

Page 7: 1007  : The Shortcut

Algebraic Rules

C). A Constant Multiplier

53 2

7 1(3 ) 4

d x y ydx x x

( )( )

nf x cxf x

c

(𝒄𝒏)𝒙𝒏−𝟏

𝑑𝑑𝑥=𝟑 [𝟓 𝒙𝟒 ]

=15

y

𝒚 ′=−𝟐𝟏 𝒙−𝟒𝒚=𝟏

𝟒 𝒙−𝟐

y’ =

Page 8: 1007  : The Shortcut

Algebraic Rules

REM:

D). A Polynomial d u v

dx

3 25 2 3 4y x x x How do you eat an elephant? One bite at a time

𝒚 ′=𝟏𝟓 𝒙𝟐+𝟒 𝒙−𝟑+𝟎

Page 9: 1007  : The Shortcut

Example: Positive Integer Powers, Multiples, Sums, and Differences

4 2 3Differentiate the polynomial 2 194

That is, find .

y x x x

dydx

Calculator: [F3] 1: d( differentiateor

[2nd ] [ 8 ] d(

d(expression,variable)

d( x^4 + 2x^2 - (3/4)x - 19 , x )

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Do it all !

36 23

43y x xx

Step 1: Rewrite using exponents

𝒚=𝟑 𝒙𝟔−𝟒 𝒙−𝟑+𝒙𝟐𝟑

Must rewrite using exponents!

𝒚 ′=𝟏𝟖 𝒙𝟓+𝟏𝟐𝒙−𝟒+𝟐𝟑 𝒙−𝟏𝟑

Page 11: 1007  : The Shortcut

A conical tank with height of 4 ft is being filled with water.

a) Write the equation for the volume of the conical tank.

b) Find the instantaneous rate of change equation of the volume with respect to the radius.

c) Find the instantaneous rate of change in Volume when the radius is 9 ft.

𝑽=𝟏𝟑 𝝅 𝒓𝟐𝒉 𝑽=

𝟏𝟑 𝝅 𝒓𝟐(𝟒) 𝑽=

𝟒𝟑 𝝅𝒓𝟐

𝑽 (𝒓 )=𝟒𝟑 𝝅 (𝒓𝟐) 𝑽 ′=𝟒

𝟑 𝝅 (𝟐𝒓 ) 𝑽 ′=𝟖𝟑 𝝅𝒓

𝑽 ′=𝟖𝟑 𝝅 (𝟗) 𝑽 ′=𝟐𝟒𝝅

When the radius is 9 the volume increases 24cu.ft.per minute

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Second and Higher Order Derivatives

2

2

The derivative is called the

of with respect to . The first derivative may itself be a differentiable function

of . If so, its derivative, ,

dyy first derivativedx

y x

dy d dy d yx ydx dx dx dx

3

3

is called the of with respect to . If double prime is differentiable, its derivative,

,is called the , and so on ...

second derivative y xy y

dy d yy third derivativedx dx

function

y’ y’’ y’’’ yiv could be y(4)

Page 13: 1007  : The Shortcut

Second and Higher Order Derivatives

(from superscript)

The multiple-prime notation begins to lose its usefulness after three primes.

So we use " super " to denote the th derivative of with respect to .

Do not c

n y nn y x

y

onfuse the notation with the

th power of , which is .

<< I like to use ROMAN NUMERALS through .>>

vi

n

nn y

y

yy

Page 14: 1007  : The Shortcut

Example: Find all the derivatives. 5 3 23 2 5 7y x x x x

𝒚 ′=𝟏𝟓 𝒙𝟒−𝟔 𝒙𝟐+𝟏𝟎 𝒙+𝟏𝒚 ′ ′=𝟔𝟎𝒙𝟑−𝟏𝟐 𝒙+𝟏𝟎𝒚 ′ ′ ′=𝟏𝟖𝟎𝒙𝟐−𝟏𝟐𝒚 𝒊𝒗=𝟑𝟔𝟎𝒙𝒚 𝒗=𝟑𝟔𝟎𝒚 𝒗𝒊=𝟎

velocity

acceleration

jerk

The rest have mathematical uses

}

Page 15: 1007  : The Shortcut

At a Joint PointPiece Wise Defined Functions:

The function must be CONTINUOUS

Derivative from the LEFT and RIGHT must be equal.

The existence of a derivative indicates a smooth curve; therefore, …

3 , 1( )

5 , 1x x

f xx x

2

2 1, 1( )

2, 1x x

f xx x

𝑑𝑦𝑑𝑥 |𝒙<𝟏=𝟐

𝑑𝑦𝑑𝑥 |𝒙 ≥𝟏=𝟐 𝒙

At 2=2 the derivative exists and it is a smooth transition

𝑑𝑦𝑑𝑥|𝒙≤𝟏=𝟏

𝑑𝑦𝑑𝑥|𝒙>−𝟏=−𝟏

Therefore the derivative DNE

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Last Update

• 08/12/10