Kevin Lee
Marc 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. . .
d
d
[ f ( x)  g ( x)]  f ' ( x)  g ' ( x)
(c )  0
dx
dx
d
dx
(cf ( x))  c  f ' ( x)
1
dx
dx
d
n
n 1
The Power Rule:
( x )  nx
dx
OKAY! Lets do
an example
y = 347,356
Uh. . .Marc. . .you
look confused!!!
What’s the matter
son!?!?
I, I, I thought. .
.why is there a
y? Shouldn’t it
be dy/dx?
How could I
forget?!?!
1) f’(x)
2) dy/dx
3) y’
4) Dx[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.
I’M READY! I’M READY!
I’M READY!
1) y = 347,356
y’= 0
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
So what
what isis the
the
So
derivative of
of
derivative
number
one?
numberthree?
two?
number
number
four?
five?
5 +4 0-3/4
y’=y’
=3(1/4)(3x)
54x
g’(x)
=zero
one
60x
Very GOOD
Marc!!! Lets see if
you can do this
one. . .
y = 3x8 – 8x3
y’ = 24x7 – 24x2
Oh that’s easy!!! It’s 24x
to the
WOOO
seventh
– HAH!!
minus
24x squared.
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!
1) d(sinx)/dx
= cosx
In this section of the
lesson there are
derivatives of
trigonometry functions
such as sine and
cosine. Lets do some
examples!
2) d(cosx)/dx
= -sinx
1) y = 5cosx
2) y = sin x + cos x
y’= cosx - sinx
dy/dx = -5sinx
3) y = x-2 + x – cosx + 3
y’= -2x-2-1+ 1 – (-sinx) + 0
y’ = (-2/x3) + sinx + 1
**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)
1) y = (4x3 – 1)2 at (3,1)
y = 16x6 – 8x3 + 1
dy/dx = 96x5 – 24x2
dy/dx at (3,1) = 96(3)5 – 24(3)2
dy/dx at (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???
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
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
**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 = (0)4 – 8(0)2 + 18
y = 18
**Points: (2,2), (-2,2), (0,18)
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)
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 ground
1) 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!
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)
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
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
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
The End