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)