by: hunter faulk anthony stephens meghan troutman

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By: Hunter Faulk Anthony Stephens Meghan Troutman

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Page 1: By: Hunter Faulk Anthony Stephens Meghan Troutman

By: Hunter FaulkAnthony StephensMeghan Troutman

Page 3: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Newton and Leibniz shared credit for the development of the differential and integral calculus. They both contributed in the math world by developing new areas of calculus such as differentiation, integration, curves, and optics. Both published their facts that included integral and derivative notation we still use today.

Page 4: By: Hunter Faulk Anthony Stephens Meghan Troutman

The Derivative of the Functions Will Use Notation That Depends on the

FunctionFunction First Derivative Second Derivative

F(x) F’(x) F’’(x)

G(x) G’(x) G’’(x)

Y Y’ or dy dx

Y’’ or

Page 5: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Definition of the Power Rule The Power Rule of Derivatives gives the following: For any real number n,the derivative of f(x) = xn is f ’(x) = nxn-1 which can also be written as

• Definition of the quotient ruleRule for finding the derivative of a quotient of two functions. If

both f and g are differentiable, then so is the quotient f(x)/g(x). In abbreviated notation, it says

(f/g) = (′ gf − ′ fg )/′ g2.

Page 6: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Definition of the product rule Rule for finding the derivative of a product of

two functions. If both f and g are differentiable, then (fg) = ′ fg + ′ f′g.

• Chain rule The formula is . Another form of

the chain rule is .

Page 8: By: Hunter Faulk Anthony Stephens Meghan Troutman

Number oney=

Number two y=

Page 9: By: Hunter Faulk Anthony Stephens Meghan Troutman

Work Page

Page 10: By: Hunter Faulk Anthony Stephens Meghan Troutman

Answers to Power Rule Problems

• Problem One

• Problem Two

Original Function is blue. Derivative is red.

Page 11: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Used when given composite functions. A composite function is a function inside another function. F(G(x))

• First you take the derivative of the outside function, while leaving the inside function alone. Then you multiply this by the derivative of the inside function, with respect to its variable x. If y= f(g(x)), then y’=

Page 13: By: Hunter Faulk Anthony Stephens Meghan Troutman

Work Page

Page 15: By: Hunter Faulk Anthony Stephens Meghan Troutman

Answer to Chain Rule Problem Two

Problem two

Page 16: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Bottom function times the derivative of the top minus the top function times the derivative of the bottom. Then divide the whole thing by the bottom function squared. If f(x)= , then f’(x)=

Page 17: By: Hunter Faulk Anthony Stephens Meghan Troutman

Quotient Rule Examples

• Problem one

• Problem two

Page 18: By: Hunter Faulk Anthony Stephens Meghan Troutman

Work Page

Page 19: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Problem one

Page 20: By: Hunter Faulk Anthony Stephens Meghan Troutman

Answer to Quotient Problem Number Two

• Problem Two

(x-2 + x-6)(-3x-4 + 8x-9)– (x-3- x-8)(-2x-3- 6x-7)(x-2 + x-6) 2

Page 21: By: Hunter Faulk Anthony Stephens Meghan Troutman

When a function involves two terms multiplied together, we use the Product Rule.

To find the derivative of two things multiplied by each other, you multiply the first function by the derivative of the second, and add that to the second function multiplied by the derivative of the first.

If F(x) = uv, then f’(x) = u

Page 23: By: Hunter Faulk Anthony Stephens Meghan Troutman

Work Page

Page 24: By: Hunter Faulk Anthony Stephens Meghan Troutman

F’(x)= (9x2+4x)(3x2-10x) + (x3-5x2)(18x+4)

Page 26: By: Hunter Faulk Anthony Stephens Meghan Troutman

• Implicit differentiation is the procedure of differentiating an implicit equation with respect to the desired variable while treating the other variables as unspecified functions of x.

• taking d of another variable

• Y2 -> 2y

Page 27: By: Hunter Faulk Anthony Stephens Meghan Troutman

Examples ofImplicit Differentiation

• Problem OneX2 + y2 = 5

• Problem TwoX2 + 3xy + y2 = ∏

Page 28: By: Hunter Faulk Anthony Stephens Meghan Troutman

Work Page

Page 30: By: Hunter Faulk Anthony Stephens Meghan Troutman

Answer to Implicit Differentiation Problem

Two

2x + 3x + 3y + 2y = 0 (3x +2y) = -2x - 3y

= -2x – 3y 3X +2Y

Page 31: By: Hunter Faulk Anthony Stephens Meghan Troutman

A method for finding the derivative of functions such as y = xsin x and

Y= lny=xln3 =x(0)+ ln3(1) =ln3 = ln3

Page 33: By: Hunter Faulk Anthony Stephens Meghan Troutman

Work Page

Page 34: By: Hunter Faulk Anthony Stephens Meghan Troutman

Y=lny=xlnx

=x( )+lnx

= (1+lnx)

Page 35: By: Hunter Faulk Anthony Stephens Meghan Troutman

Answers to Logarithmic Differentiation Problem

TwoY=

=sinx + lnxcosx= ( +lnx(cosx))

www.schools.sd68.bc.ca

Page 36: By: Hunter Faulk Anthony Stephens Meghan Troutman

1978 AB 2 Let f(x) = (1 - x)2 for all real numbers x, and let

g(x) = ln(x) for all x > 0. Let h(x) = ( 1 - ln(x))2. a. Determine whether h(x) is the composition

f(g(x)) or the composition g(f(x)). b. Find h (x). ′c. Find h″(x). d. On the axes provided, sketch the graph of h.

Page 37: By: Hunter Faulk Anthony Stephens Meghan Troutman

FRQ Answer

A) f(g(x)) = f(lnx) = (1-lnx)2g(f(x)) = g((1-x)2) = ln((1-x)2)

Therefore h(x) = f(g(x))B) h’(x) = 2(1-lnx)(1/x) = 2 lnx-1

C) h’’(x) = 2 x(1/x) – (lnx-1) x2

= 2 * 2 – lnx x2D) The Inflection point is at (-e2, 1)

The minimum is at (e, 0 )

Page 38: By: Hunter Faulk Anthony Stephens Meghan Troutman

FRQ Example

1977 AB 7 BC 6 Let f be the real-valued function defined by f(x)

= sin3(x) + sin3|x|. a. Find f (x) for x > 0. ′b. Find f’(x) for x < 0. c. Determine whether f(x) is continuous at x =

0. Justify your answer. d. Determine whether the derivative of f(x)

exists at x = 0. Justify your answer.

Page 39: By: Hunter Faulk Anthony Stephens Meghan Troutman

FRQ Answer

A) For x > 0F(x) = sin3x + sin3x = 2sin3xF’(x) = 6sin2xcosx

B) For x > 0f(x) = sin3x + sin3(-x) = sin3x - sin3x = 0F’(x) = 0

C) f(0) = 0Lim f(x) = lim 2sin3x = 0Lim f(x) = lim 0 = 0

Since Lim f(x) = 0 = f(0), the function of f is continous at x = 0D) F’(x) = lim f(x+h) – f(x) if the limit exists hAt x = 0Lim f(h) - f(0) = 0 hTherefore, lim f(h) - f(0) and so f’(0) exists and equals 0. h

X ->0+X ->0+

X ->0- X ->0+

X ->0+

H -> 0

H -> 0

H -> 0

Page 40: By: Hunter Faulk Anthony Stephens Meghan Troutman

Bibliography• http://www.mathwords.com/f/formula.htm• http://images.google.com/imgres?imgurl=http://www.francis.edu/

uploadedImages/Math/blackboard_math.gif&imgreful• www.open.salon.com• www.babble.com • www. school-clipart.com• www.first90days.wordpress.com• www.schools.sd68.bc.ca • http://numericalmethods.eng.usf.edu/anecdotes/newton.html• http://gardenofpraise.com/ibdnewt.htm• http://scienceworld.wolfram.com/biography/leibniz/html• http://calculusthemusical.com/wp-content/uploads/2008/02/matheatre-power-

rule.mp3©Copyright. Hunter Faulk. Megan Troutman. Anthony Stevens. February 19, 2010