Lesson+Summaries+-+Derivatives

Introduction to Derivatives February 16, 2011 Nicole Saunders

WHAT IS A DERIVATIVE?
 * Slope of a tangent at a point
 * instantaneous rate of change at that point
 * the limit as //x// approaches zero [f(a+h)-f(a)]/h
 * y’ or dy/dx

http://www.youtube.com/watch?v=HOHEFE_sIl8

The derivative of a function y = f(x) at the point (x,f(x)) equals the gradient of the tangent line to the graph at that point. It can be defined as:

where 'h' approaches zero as a limit. This diagram ill ustrates this concept graphically: The derivative formula (above) gives the gradient of the secant line between the two points. As the value of 'h' gets smaller, the two points get closer and the gradient of the secant approaches that of the tangent line to the curve at (x,f(x)): < [|__http://www.teacherschoice.com.au/first_principles.htm__] >.

EXAMPLES: 1.) Use the first principles to determine the derivative for each function and solve f’(1) f(x)=x^2 <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) [(x+h)^2 - (x)^2]/h <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; margin: 0px 0px 16px; padding: 0px;"><span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) [x^2 +2hx + h^2 - x^2] /h <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; margin: 0px 0px 16px;"><span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) [h(2 x+h)] /h <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; margin: 0px 0px 16px; padding: 0px;"> f’(x)= lim (h <span style="font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial;">→0) 2x + h <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px 0px 16px; padding: 0px;">f’(x)= 2x <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px 0px 16px; padding: 0px;">2.) Use the first principles to determine the derivative for each function and solve f’(1) <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px 0px 16px; padding: 0px;">f(x)=x^3 <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; margin: 0px 0px 16px; padding: 0px;"><span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) [(x+h)^3 - (x)^3] / h <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; margin: 0px 0px 16px; padding: 0px;"><span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) [((x+h)-x)(x+h)^2+(x+h)+x^2)]/h <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; margin: 0px 0px 16px; padding: 0px;"><span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) [(h)(x^2 + 2xh +h^2 + x^2 +hx +x^2)]/h <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; margin: 0px 0px 16px; padding: 0px;"><span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial;">f’(x)= lim (h →0) 3x^2 + 3xh +h^2 <span style="color: #333233; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; letter-spacing: 0px; margin: 0px 0px 16px; padding: 0px;">f’(x)= 3x^2 <span style="color: #b70f0a; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; letter-spacing: 0px; margin: 0px 0px 16px; padding: 0px;">Have a look at this page for more information: <span style="background-position: 100% 50%; cursor: pointer; font-size-adjust: none; font-stretch: normal; font: 13px/normal Arial; letter-spacing: 0px; padding-right: 10px;">[|__http://www.mathsisfun.com/calculus/derivatives-introduction.html__] <span style="color: #000000; font-size-adjust: none; font-stretch: normal; font: 13px/19px Arial; letter-spacing: 0px; margin: 0px 0px 16px; padding: 0px;">media type="youtube" key="HOHEFE_sIl8" height="390" width="640"

= **How to Factor a Sum or Difference of Cubes:** = =<span style="font-family: "Times New Roman"; font-size: 12pt; margin: 0px; padding: 5px 0px 0px; position: relative; top: 16pt;"> =

<span style="font-family: "Times New Roman",serif; font-size: 14pt; line-height: 21px; margin: 0cm 0cm 10pt; padding: 0px;">Notes (February 18th, 2011) – Ajmeet Dhillon __<span style="font-family: "Times New Roman",serif; font-size: 12pt;">SOLVING DERIVATIVES __ <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt; padding: 0px;">Solving Derivatives by the Ist Principle can be a bit painful.

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt; padding: 0px;">Shortcuts <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px; text-indent: -18pt;">1) Power Rule <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px; text-indent: -18pt;">- Using Ist Principle to find the derivative of f(x) = xn where n€N. <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim (x+h)n - xn <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0 h

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">hint: an – bn = (a-b)(an-1+an-2b+an-3b2+.....a2bn-3+abn-2+bn-1)

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim (x+h)n - xn <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0 h

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim [(x+h)-x][(x+h)n-1+(x+h)n-2x+(x+h)n-3x2+.....(x+h)2xn-3+(x+3)xn-2+xn-1] <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0 h <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">we will cancel out the common terms at this step.

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">After cancelling, we get <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim [(x+h)n-1+(x+h)n-2x+(x+h)n-3x2+.....(x+h)2xn-3+(x+3)xn-2+xn-1] <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">After applying the h→0, we get rid of limit. <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">= xn-1+xn-2x+xn-3x+.......x2xn-3+x.xn-2+xn-1 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">= xn-1+xn-1+xn-1...........xn-1+xn-1+xn-1 (we use exponential sum rule to get n-1 as power <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">over each term). <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">= nxn-1 (since we have n terms).

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">Therefore if f(x) = xn n€W <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">then f’x = nxn-1

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt; padding: 0px;">Examples: <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt; padding: 0px;">Differentiate: <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt; padding: 0px;">1) f(x) = x5 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt; padding: 0px;">f’(x) = 5x4

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px; text-indent: -18pt;">2) f(x) = (1/2)x7 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">f’(x) = (7/2)x6

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px; text-indent: -18pt;">3) f(x) = 2x3 – 4x2 + 5 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">f’(x) = 6x2 – 8x

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px; text-indent: -18pt;">4) f(x) = 1/x <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">f’(x) = x-1 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">= -x-2 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">= -(1/x2)

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px; text-indent: -18pt;">5) 53√x <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">= 5x(1/3) <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">= (5/3)x(-2/3) <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">= 5/3x(2/3) <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">= 5/3 3√x

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">Homework - Finish the handout on Practice with Product Rule and work on the Different Kinds of Numbers Assignment. <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt; padding: 0px;">media type="youtube" key="fUVf-qKaGzo" height="390" width="480" <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 36pt;">Here is a video that covers the concepts similar to the one discussed in our notes.

= **The Product Rule** = Tuesday, February 22, 2011, by Joel Tham

When taking the derivative of the product of two functions: For example:

To find the derivative of the product of two functions, use the product rule:

**This video explains the product rule:**
(only watch the first 5 minutes as the chain rule is included after that)

media type="youtube" key="h78GdGiRmpM" height="390" width="480" align="center"

**Homework:** Homework sheet Independent Quick Study Assignment: There are Different Kinds of Numbers?????? Teamwork Tuesday Assignment


 * February 23, 2011**
 * Rabeea Fatima**
 * AMC math contest. No math class.**

= Finding The Second Derivative = Friday, February 25, 2011, by Shen Wang

To find the second derivative. We take the derivative twice. f (x) -> f '(x) -> f ''(x)

Solving Motion Problems:

Definition Reminders:

Displacement -> Change in position with time. Velocity -> Rate of change in displacement over a certain period of time. (First Derivative of Displacement) Acceleration -> Rate of change in velocity over a certain period of time. (Second Derivative of Displacement or First Derivative of Velocity)

Ex) A Baseball is hit vertically upwards. The height, s, in metres, of the ball after t seconds can be modelled by s(t) = -5t^2 + 30t + 1, t must be greater than or equal to 0.

a) Determine the average velocity of the baseball between 4 and 6 seconds.

avgV = [s(6) - s(4)]/(6-4)

b) Determine velocity of the ball at 4 & 6 seconds s'(4) =  s'(6) =

c) When will the ball hit the ground? s(t) = 0

d) What is the balls impact velocity? s'(final time) =

e) Determine the ball's acceleration. s''(t)

f) What is the max height of the ball? s'(t) = 0

Homework:
Do Velocity handout.

=__ INTRO TO THE CHAIN RULE __= Mon, Feb. 28, 2011 1234567 Mengdi Wei

__Example.__
 * Recall //Composition Functions//** asdfasdf g o f = g( f(x) )

If f(x) and g(x) are differentiable, the derivative of the composition function h(x) = f(g(x)) is:
 * h'(x) = f'(g(x)) · g'(x) **
 * ** i.e. find the derivative of the outside function as a whole using power rule, and multiply by the derivative of the inside function

__Example__ Differentiate using the chain rule:

For some people, the Leibniz Chain Rule notation is easier to understand since you can kinda treat the it as a fraction calculation. Here's a video showing how you could end up with what you would get from the h'(x) = f'(g(x)) · g'(x) chain rule, by using the Leibniz notation. media type="youtube" key="C5aEbEPt_ig" height="390" width="480"[]

When dealing with more complicated question involving chain rule, you could start from finding the derivative of the most inside equation/function, and then work your way out by multiplying the derivative of the outside equation/function found by using the power rule. Here's an example (go to 4:40 for the more complicated example) media type="youtube" key="6_lmiPDedsY" height="390" width="480"﻿[]

//**And Something Extra for those who are interested...**// Here's a video on the General Chain Rule. It is similar to the Leibniz one but could involve more than one "u"-kind functions. media type="youtube" key="HOYA0-pOHsg" height="390" width="480" [] (Part 1, you could find Part 2 in the link list on the right)

__**Homework:**__ Chain Rule Practice (hand out, required to do every single question!)

Jane Yang March 1, 2011
 * The Quotient Rule **



__Deriving the Quotient Rule (handout)__

-used for rational functions, ,
 * Summary**

//Examples://

1.

//2.//

// The Quotient Rule video // - there is a **mnemonic** in the video if you have trouble remembering the rule media type="youtube" key="K3MxofAF-9o" height="390" width="480" []

//** Homework **// - finish the investigation for the quotient rule (Part A and B) - finish the reverse side of this sheet ("Quotient Rule Homework")


 * March 2, 2011**
 * Virginia Lee**
 * Topic: None**

Today we did not have class because of our trip to the Nanotech Lab the University of Toronto. We will do some review on Friday and we will also have a work period this coming Monday.

If you want to do some more practice or review, this website has some good examples (press the book or the pencil icon): []

Also, here's an interesting song about derivatives and its applications:

media type="youtube" key="eEhkBmHqGA8" height="312" width="384" align="left"

[]


 * Homework:** Remember to print the practice test from the wiki and begin studying for our test on Tuesday March 8.

Nicole Saunders March 5, 2011 Review

__//**What is a derivative?**//__


 * <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px; padding: 0px 0px 0px 3em;">Slope of a tangent at a point
 * <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px; padding: 0px 0px 0px 3em;">instantaneous rate of change at that point
 * <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px; padding: 0px 0px 0px 3em;">the limit as //x// approaches zero [f(a+h)-f(a)]/h
 * <span style="font-size-adjust: none; font-stretch: normal; font: 12px/normal Arial; letter-spacing: 0px; margin: 0px; padding: 0px 0px 0px 3em;">y’ or dy/dx

How can you find the derivative?


 * //__1.) THE POWER RULE__//**

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px; text-indent: -18pt;">- Using Ist Principle to find the derivative of f(x) = xn where n€N. <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim (x+h)n - xn <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0 h

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">hint: an – bn = (a-b)(an-1+an-2b+an-3b2+.....a2bn-3+abn-2+bn-1)

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim (x+h)n - xn <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0 h

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim [(x+h)-x][(x+h)n-1+(x+h)n-2x+(x+h)n-3x2+.....(x+h)2xn-3+(x+3)xn-2+xn-1] <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0 h <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">we will cancel out the common terms at this step.

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">After cancelling, we get <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">f’(x) = lim [(x+h)n-1+(x+h)n-2x+(x+h)n-3x2+.....(x+h)2xn-3+(x+3)xn-2+xn-1] <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">h→0

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">After applying the h→0, we get rid of limit. <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">= xn-1+xn-2x+xn-3x+.......x2xn-3+x.xn-2+xn-1 <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">= xn-1+xn-1+xn-1...........xn-1+xn-1+xn-1 (we use exponential sum rule to get n-1 as power <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">over each term). <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">= nxn-1 (since we have n terms).

<span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">Therefore if f(x) = xn n€W <span style="font-family: "Times New Roman",serif; font-size: 12pt; line-height: normal; margin: 0cm 0cm 0pt 54pt; padding: 0px;">then f’x = nxn-1

__//**2.) THE PRODUCT RULE**//__

When taking the derivative of the product of two functions: For example: **//__ 3.) THE CHAIN RULE __//** used when functions are composed within one another  h(x)=f'(g(x)) x g(x)

__//**4.) THE QUOTIENT RULE**//__ used for fractions media type="youtube" key="ANyVpMS3HL4" height="390" width="640" media type="youtube" key="IePCHjMeFkE" height="390" width="640" media type="youtube" key="z1lwai-lIzY" height="390" width="480" HOMEWORK STUDY!

Unit Test: Rates of Change and Derivatives Basic March 8, 2011 by Rosetta D'souza Homework: N/A Just for fun: Pasted from <[]>