AP Calculus AB Final Project 2.2 Basic Differentiation Rule

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AP Calculus AB Final Project2.2 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 LeeMarc Reynaud Kayla Toomer

Period 1 & 13

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!!!

LOOK ALIVE MARC!!!Anyways. . .

The Power Rule:

Formulas

OKAY! Lets do an example

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?

How could I forget?!?!

Examples of Differentiation1) f(x)2) dy/dx3) y4) Dx[y]

Here are the different forms of dy/dx but they can also come in different variables. Now lets do some examples.5) dh/dtIM READY! IM READY! IM READY!

1) y = 347,356y= 02) y = xy = 13) y = 12x5 y = 60x44) y = 9x6 + 4y = 54x5 + 0

So what is the derivative of number one?zero5) g(x) = (3x)1/4 g(x) = (1/4)(3x)-3/4g(x) = 3(1/4)(3x)-3/4oneSo what is the derivative of number two? y = 60x4So what is the derivative of number three? y = 54x5 + 0So what is the derivative of number four?g(x) = 3(1/4)(3x)-3/4So what is the derivative of number five?

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

y = 3x8 8x3Oh thats easy!!! Its 24x to the seventh minus 24x squared. y = 24x7 24x2

WOOO HAH!!

WHYYYYYY?!?!?!?!SIMPLIFY, SIMPLIFY, SIMPLIFY!!!!

y = 24x7 24x2y = 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!

First Derivative ofSine & Cosine Functions1) d(sinx)/dx2) 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!

Example Problems1) y = 5cosx2) y = sin x + cos xdy/dx = -5sinx 3) y = x-2 + x cosx + 3 y = (-2/x3) + sinx + 1y= cosx - sinxy= -2x-2-1+ 1 (-sinx) + 0

Steps to Finding theDerivative1) Change radicals to exponential form2) Simplify expression by multiplying or dividing similar bases

3) Remove variable from the denominator4) Now apply the power rule5) Make sure you call your answer the derivative of the function (y dy/dx)**Simplify expression before finding derivative

Finding DerivativeAt a Certain Point1) y = (4x3 1)2 at (3,1)y = 16x6 8x3 + 1dy/dx = 96x5 24x2dy/dx at (3,1) = 96(3)5 24(3)2dy/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.

Steps to Determine aHorizontal Tangent Line1) Take derivative2) Set derivative = 0

3) Algebraically find the points4) After finding x or the root re-enter the value into the original equation to find the value of y

Lets do an example

Example Problem1) Determine points where function has horizontal tangent line:y = x4 8x2 + 18y = 4x3 16x = 0 m = y = 00 = 4x(x2 4)x = 2, 0

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

WARNING

WHAT YOU ARE ABOUT TO READ IS VERY VERY VERY VERY IMPORTANT!

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)

Free Falling Objects and Velocity Problem

Let s represent the height off the ground of a free falling object. Then s(t) = -16t2 + v0t + s0 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 ground1) Write an equation for s(t).v0 = 0 ; s0 = 100 ft s(t) = -16t2 + 100

Well at least we will be able to find the velocity of the falling pieces of the sky. HAHA!

vavg = -32 ft/sec = s(t)m = s(t) = v(t) = vFree Falling Objects and Velocity Problem2) Find the average velocity for the first 2 seconds (t = 0 t = 2).vavg = dx/dt = [s(2) s(0)] / (2-0)

v(t) = s(2) = -64 ft/secv(t) = s(t) = -32tFree Falling Objects and Velocity Problem3) Find the instantaneous velocity at t = 2 (v(t) or s(t)).

t = 10/4 sec = 5/2 secs(t) = 0Free Falling Objects and Velocity Problem4) How many seconds will it take to reach the ground? (When s(t) = 0).

0 = -16t2 + 100(t2)1/2 = (100/16)1/2

v(5/2) = -80 ft/secs(5/2) = v(5/2)Free Falling Objects and Velocity Problem5) 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)

The End