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AP Calculus AB Final Project 2.2 Basic Differentiation Rule. Objective/SWBAT : Use the basic differentiation method in order to find the derivative of a function without using the definition of the derivative. Kevin Lee Marc Reynaud Kayla Toomer Period 1 & 13. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
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1. AP Calculus AB Final Project 2.2 Basic Differentiation Rule

2. Objective/SWBAT: Use the basic differentiation method in order to find the derivative of a function without using the definition of the derivative. Kevin Lee Marc Reynaud Kayla Toomer Period 1 & 13

3. Hello class!!! Welcome to AP Calculus AB, taught by Mr. Spitz. . .which is ME! Today we are learning about Basic Differentiation a.k.a GARBAGE!!!

4. LOOK ALIVE MARC!!! Anyways. . .

5. Formulas The Power Rule:

6. OKAY! Lets do an example

7. y = 347,356 I, I, I thought. . .why is there a y? Shouldn’t it be dy/dx? Uh. . .Marc. . .you look confused!!! What’s the matter son!?!?

8. How could I forget?!?!

9. Examples of Differentiation 1) f’(x) 2) dy/dx 4) Dx[y] 3) y’ 5) dh/dt Here are the different forms of dy/dx but they can also come in different variables. Now lets do some examples.

11. y’= 0 1) y = 347,356 2) y = x y’ = 1 3) y = 12x5 y’ = 60x4 4) y = 9x6 + 4 y’ = 54x5 + 0 5) g(x) = (3x)1/4 g’(x) = (1/4)(3x)-3/4 g’(x) = 3(1/4)(3x)-3/4 y’ = 54x5 + 0 g’(x) = 3(1/4)(3x)-3/4 zero one y’ = 60x4 So what is the derivative of number three? So what is the derivative of number one? So what is the derivative of number two? So what is the derivative of number four? So what is the derivative of number five?

12. Very GOOD Marc!!! Lets see if you can do this one. . .

13. y = 3x8 – 8x3 y’ = 24x7 – 24x2 Oh that’s easy!!! It’s 24x to the seventh minus 24x squared. WOOO – HAH!!

14. WHYYYYYY?!?!?!?! SIMPLIFY, SIMPLIFY, SIMPLIFY!!!!

15. y’ = 24x7 – 24x2 y’ = 24x2(x5 – 1) Yes the answer that you have given is only half-way correct. You have to simplify that jon! YEA. . . I understand now Spity Cent!

16. First Derivative ofSine & Cosine Functions 2) d(cosx)/dx 1) d(sinx)/dx = cosx = -sinx In this section of the lesson there are derivatives of trigonometry functions such as sine and cosine. Lets do some examples!

17. Example Problems 1) y = 5cosx 2) y = sin x + cos x dy/dx = -5sinx y’= cosx - sinx 3) y = x-2 + x – cosx + 3 y’= -2x-2-1+ 1 – (-sinx) + 0 y’ = (-2/x3) + sinx + 1

18. Steps to Finding theDerivative **Simplify expression before finding derivative 1) Change radicals to exponential form 2) Simplify expression by multiplying or dividing similar bases 3) Remove variable from the denominator 4) Now apply the power rule 5) Make sure you call your answer the derivative of the function (y  dy/dx)

19. Finding DerivativeAt a Certain Point 1) y = (4x3 – 1)2 at (3,1) y = 16x6 – 8x3 + 1 dy/dx = 96x5 – 24x2 dy/dxat (3,1) = 96(3)5 – 24(3)2 dy/dxat (3,1) = 23,112 Still follow the same steps that you would normally do to find the derivative. The only twist is that the coordinate point that is given needs to be inserted in the derivitized formula to find the value of the derivitization at that point. How do I do this one???

20. Steps to Determine a Horizontal Tangent Line 1) Take derivative Lets do an example 2) Set derivative = 0 3) Algebraically find the points 4) After finding x or the root re-enter the value into the original equation to find the value of y

21. Example Problem 1) Determine points where function has horizontal tangent line: y = x4 – 8x2 + 18 y’ = 4x3 – 16x = 0  m = y’ = 0 0 = 4x(x2 – 4) x = ± 2, 0

22. Example Problem **Plug x values into original equation to solve for y and get the points where the function has a horizontal tangent line. y = (± 2)4 – 8(± 2)2 + 18 y = 2 y = 18 y = (0)4 – 8(0)2 + 18 **Points: (2,2), (-2,2), (0,18)

23. WARNING WHAT YOU ARE ABOUT TO READ IS VERY VERY VERY VERY IMPORTANT!

24. An Important Theorem Differentiability at a point implies continuity at a point. However, continuity at a point does not imply differentiability at that point (only works one way). **If f is differentiable at point x=c then f is continuous at x=c (f’(x) exists)

25. Free Falling Objects and Velocity Problem Well at least we will be able to find the velocity of the falling pieces of the sky. HAHA! Let s represent the height off the ground of a free falling object. Then s(t) = -16t2 + v0t + s0where “t” is in seconds and “s” is in feet. *Example Problem: Consider a rock dropped off the end of a cliff 100 feet above the ground 1) Write an equation for s(t). v0 = 0 ; s0 = 100 ft s(t) = -16t2 + 100

26. Free Falling Objects and Velocity Problem 2) Find the average velocity for the first 2 seconds (t = 0  t = 2). m = s’(t) = v(t) = v vavg= dx/dt = [s(2) – s(0)] / (2-0) vavg = -32 ft/sec = s’(t)

27. Free Falling Objects and Velocity Problem 3) Find the instantaneous velocity at t = 2 (v(t) or s’(t)). v(t) = s’(t) = -32t v(t) = s’(2) = -64 ft/sec

28. Free Falling Objects and Velocity Problem 4) How many seconds will it take to reach the ground? (When s(t) = 0). s(t) = 0 0 = -16t2 + 100 (t2)1/2 = (100/16)1/2 t = 10/4 sec = 5/2 sec

29. Free Falling Objects and Velocity Problem 5) What is the velocity of the rock just before it hits the ground? (Use t value from above problem). s’(5/2) = v(5/2) v(5/2) = -32(5/2) v(5/2) = -80 ft/sec

30. The End