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 illustrates 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 f’(x)= lim (h→0) [(x+h)^2 - (x)^2]/h f’(x)= lim (h→0) [x^2 +2hx + h^2 - x^2] /h f’(x)= lim (h→0) [h(2x+h)] /h f’(x)= lim (h→0) 2x + h f’(x)= 2x 2.) Use the first principles to determine the derivative for each function and solve f’(1) f(x)=x^3 f’(x)= lim (h→0) [(x+h)^3 - (x)^3] / h f’(x)= lim (h→0) [((x+h)-x)(x+h)^2+(x+h)+x^2)]/h f’(x)= lim (h→0) [(h)(x^2 + 2xh +h^2 + x^2 +hx +x^2)]/h f’(x)= lim (h→0) 3x^2 + 3xh +h^2 f’(x)= 3x^2 Have a look at this page for more information: __http://www.mathsisfun.com/calculus/derivatives-introduction.html__
Notes (February 18th, 2011) – Ajmeet Dhillon SOLVING DERIVATIVES Solving Derivatives by the Ist Principle can be a bit painful.
Shortcuts 1) Power Rule - Using Ist Principle to find the derivative of f(x) = xn where n€N. f’(x) = lim (x+h)n - xn h→0 h
hint: an – bn = (a-b)(an-1+an-2b+an-3b2+.....a2bn-3+abn-2+bn-1)
f’(x) = lim (x+h)n - xn h→0 h
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] h→0 h we will cancel out the common terms at this step.
After cancelling, we get 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] h→0
After applying the h→0, we get rid of limit. = xn-1+xn-2x+xn-3x+.......x2xn-3+x.xn-2+xn-1 = xn-1+xn-1+xn-1...........xn-1+xn-1+xn-1 (we use exponential sum rule to get n-1 as power over each term). = nxn-1 (since we have n terms).
Homework - Finish the handout on Practice with Product Rule and work on the Different Kinds of Numbers Assignment.
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)
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
Recall Composition Functionsasdfasdfg o f = g( f(x) ) Example.
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. http://www.youtube.com/watch?v=C5aEbEPt_ig
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)
http://www.youtube.com/watch?v=6_lmiPDedsY
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.
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.
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?
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
How can you find the derivative?
1.) THE POWER RULE
- Using Ist Principle to find the derivative of f(x) = xn where n€N. f’(x) = lim (x+h)n - xn h→0 h
hint: an – bn = (a-b)(an-1+an-2b+an-3b2+.....a2bn-3+abn-2+bn-1)
f’(x) = lim (x+h)n - xn h→0 h
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] h→0 h we will cancel out the common terms at this step.
After cancelling, we get 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] h→0
After applying the h→0, we get rid of limit. = xn-1+xn-2x+xn-3x+.......x2xn-3+x.xn-2+xn-1 = xn-1+xn-1+xn-1...........xn-1+xn-1+xn-1 (we use exponential sum rule to get n-1 as power over each term). = nxn-1 (since we have n terms).
Therefore if f(x) = xn n€W 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)
February 16, 2011
Nicole Saunders
WHAT IS A DERIVATIVE?
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 illustrates 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
f’(x)= lim (h→0) [(x+h)^2 - (x)^2]/h
f’(x)= lim (h→0) [x^2 +2hx + h^2 - x^2] /h
f’(x)= lim (h→0) [h(2x+h)] /h
f’(x)= lim (h→0) 2x + h
f’(x)= 2x
2.) Use the first principles to determine the derivative for each function and solve f’(1)
f(x)=x^3
f’(x)= lim (h→0) [(x+h)^3 - (x)^3] / h
f’(x)= lim (h→0) [((x+h)-x)(x+h)^2+(x+h)+x^2)]/h
f’(x)= lim (h→0) [(h)(x^2 + 2xh +h^2 + x^2 +hx +x^2)]/h
f’(x)= lim (h→0) 3x^2 + 3xh +h^2
f’(x)= 3x^2
Have a look at this page for more information: __http://www.mathsisfun.com/calculus/derivatives-introduction.html__
How to Factor a Sum or Difference of Cubes:
Notes (February 18th, 2011) – Ajmeet Dhillon
SOLVING DERIVATIVES
Solving Derivatives by the Ist Principle can be a bit painful.
Shortcuts
1) Power Rule
- Using Ist Principle to find the derivative of f(x) = xn where n€N.
f’(x) = lim (x+h)n - xn
h→0 h
hint: an – bn = (a-b)(an-1+an-2b+an-3b2+.....a2bn-3+abn-2+bn-1)
f’(x) = lim (x+h)n - xn
h→0 h
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]
h→0 h
we will cancel out the common terms at this step.
After cancelling, we get
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]
h→0
After applying the h→0, we get rid of limit.
= xn-1+xn-2x+xn-3x+.......x2xn-3+x.xn-2+xn-1
= xn-1+xn-1+xn-1...........xn-1+xn-1+xn-1 (we use exponential sum rule to get n-1 as power
over each term).
= nxn-1 (since we have n terms).
Therefore if f(x) = xn n€W
then f’x = nxn-1
Examples:
Differentiate:
1) f(x) = x5
f’(x) = 5x4
2) f(x) = (1/2)x7
f’(x) = (7/2)x6
3) f(x) = 2x3 – 4x2 + 5
f’(x) = 6x2 – 8x
4) f(x) = 1/x
f’(x) = x-1
= -x-2
= -(1/x2)
5) 53√x
= 5x(1/3)
= (5/3)x(-2/3)
= 5/3x(2/3)
= 5/3 3√x
Homework - Finish the handout on Practice with Product Rule and work on the Different Kinds of Numbers Assignment.
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 ThamWhen 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)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 WangTo 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 WeiRecall Composition Functions asdfasdfg o f = g( f(x) )
Example.
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.
http://www.youtube.com/watch?v=C5aEbEPt_ig
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)
http://www.youtube.com/watch?v=6_lmiPDedsY
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.
http://www.youtube.com/watch?v=HOYA0-pOHsg&feature=relmfu (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!)
The Quotient Rule
Jane Yang
March 1, 2011
Deriving the Quotient Rule (handout)
Summary
-used for rational functions,
Examples:
1.
2.
The Quotient Rule video
- there is a mnemonic in the video if you have trouble remembering the rule
http://www.youtube.com/watch?v=K3MxofAF-9o
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): http://www.woodbridgecollege.yrdsb.edu.on.ca/courses/calculus/
Also, here's an interesting song about derivatives and its applications:
http://www.youtube.com/watch?v=eEhkBmHqGA8
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?
How can you find the derivative?
1.) THE POWER RULE
- Using Ist Principle to find the derivative of f(x) = xn where n€N.
f’(x) = lim (x+h)n - xn
h→0 h
hint: an – bn = (a-b)(an-1+an-2b+an-3b2+.....a2bn-3+abn-2+bn-1)
f’(x) = lim (x+h)n - xn
h→0 h
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]
h→0 h
we will cancel out the common terms at this step.
After cancelling, we get
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]
h→0
After applying the h→0, we get rid of limit.
= xn-1+xn-2x+xn-3x+.......x2xn-3+x.xn-2+xn-1
= xn-1+xn-1+xn-1...........xn-1+xn-1+xn-1 (we use exponential sum rule to get n-1 as power
over each term).
= nxn-1 (since we have n terms).
Therefore if f(x) = xn n€W
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
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 <http://www.charlottemathtutor.com/htmlfile/cartoons.html>