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Kevin Lee Marc Reynaud Kayla Toomer Period 1 & 13

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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!!!

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LOOK ALIVE MARC!!! Anyways...

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The Power Rule:

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OKAY! Lets do an example

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y = 347,356 Uh...Marc...you look confused!!! Whats the matter son!?!? I, I, I thought...why is there a y? Shouldnt it be dy/dx?

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How could I forget?!?!

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

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IM READY! IM READY! IM READY!

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1) y = 347,356 y= 0 2) y = x y = 1 3) y = 12x 5 y = 60 x 4 4) y = 9x 6 + 4 y = 54 x 5 + 0 So what is the derivative of number one? zero 5) g(x) = (3x) 1/4 g(x) = (1/4)(3x) -3/4 g(x) = 3(1/4)(3x) -3/4 one So what is the derivative of number two? y = 60x 4 So what is the derivative of number three? y = 54x 5 + 0 So what is the derivative of number four? g(x) = 3(1/4)(3x) -3/4 So what is the derivative of number five?

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

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y = 3x 8 8x 3 Oh thats easy!!! Its 24x to the seventh minus 24x squared. y = 24x 7 24x 2 WOOO HAH!!

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y = 24x 7 24x 2 y = 24x 2 (x 5 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!

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

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1) y = 5cosx 2) y = sin x + cos x dy/dx = -5sinx 3) y = x -2 + x cosx + 3 y = (-2/ x 3 ) + sinx + 1 y= cosx - sinx y= -2x -2-1 + 1 (-sinx) + 0

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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) **Simplify expression before finding derivative

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1) y = (4x 3 1) 2 at (3,1) y = 16x 6 8x 3 + 1 dy/dx = 96x 5 24x 2 dy/dx at (3,1) = 96(3) 5 24(3) 2 dy/dx at (3,1) = 23,112 How do I do this one??? 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.

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1) Take derivative 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 Lets do an example

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1) Determine points where function has horizontal tangent line: y = x 4 8x 2 + 18 y = 4x 3 16x = 0 m = y = 0 0 = 4x(x 2 4) x = 2, 0

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**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 = (0) 4 8(0) 2 + 18 **Points: (2,2), (-2,2), (0,18) y = 2 y = 18

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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)

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Let s represent the height off the ground of a free falling object. Then s(t) = -16t 2 + v 0 t + s 0 where 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). v 0 = 0 ; s 0 = 100 ft s(t) = -16t 2 + 100 Well at least we will be able to find the velocity of the falling pieces of the sky. HAHA!

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v avg = -32 ft/sec = s(t) m = s(t) = v(t) = v 2) Find the average velocity for the first 2 seconds (t = 0 t = 2). v avg = dx/dt = [s(2) s(0)] / (2-0)

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v(t) = s(2) = -64 ft/sec v(t) = s(t) = -32t 3) Find the instantaneous velocity at t = 2 (v(t) or s(t)).

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t = 10/4 sec = 5/2 sec s(t) = 0 4) How many seconds will it take to reach the ground? (When s(t) = 0). 0 = -16t 2 + 100 (t 2 ) 1/2 = (100/16) 1/2

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v(5/2) = -80 ft/sec s(5/2) = v(5/2) 5) What is the velocity of the rock just before it hits the ground? (Use t value from above problem). v(5/2) = -32(5/2)

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The End