# Unit 2 Derivatives

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```					Derivatives 2.2
St. Pius X High School
Ms. Hernandez
AP Calculus I
F06 Q1 Derivatives Unit
Some Differentiation Rules!
Yeah, we have some rules that make
finding the derivative so much EASIER!
   Constant
   Power
   Constant Multiple
   Sum and Difference
   Sine and Cosine
d
[c ]  0
Constant              dx

d            d
[ y  7]  [7]  0
dx           dx
d                  d
[ f ( x )  0]  [0]  0
dx                 dx
d                 d
[ s(t )  3]  [3]  0
dt                dt
y  k 2   k is a constant    y'  0
d n
[ x ]  nx n 1
Power               dx

d
[ f ( x )  x 3 ]  3x 2
dx
2 / 3     1
g ( x)  x  x
3        1/ 3
 g '( x )  1/ 3( x )             2/3
3x
1   dy d 2         3   2
y 2     [ x ]  2 x   3
x    dx dx               x
d
[ f ( x )  x ]  ( x1 ) '  (1) x11  (1) x 0  (1)1  1
dx
Special case n=1  f(x)=x  f’(x)=1
d
[cf ( x )]  cf '( x )
Constant Multiple                       dx

d
[ y  2 x ]  2(1)  2
dx
d       7                         2 7
[ y  ]  7 f '( x )  7( 1) x  2
dx      x                            x

d             4t 3    4 3      4 2
[ f (t )       ]  (t ) '  (3t )  4t 2
dx             3      3        3

d        1  1 d 2 / 3 1                  5/ 3       1
 y  3 2   2 dx [ x ]  2 ( 2 / 3) x
dx 

2 x                                         3 3 x5
Sum and                d
[ f ( x )  g ( x )]  f '( x )  g '( x )
Difference             dx

1                                     1
f ( x)  x  2 x  x  7
4       3
f '( x )  4 x  6 x   0
3        2

5                                     5

3                                         2
7x                                    21x
f ( x)       4 x2  2 x  3      f '( x )        8x  2  0
5                                      5
Sine               and      Cosine
d                            d
[sin x ]  cos x             [cos x ]   sin x
dx                           dx

y  2 cos x  y '  2sin x
sin x         cos x
y         y' 
2             2
y  x  cos x  3sin x  y '  1  sin x  3cos x 
TS 2 Rates of 
PVA = Position, Velocity, & Acceleration
RATE OF CHANGE
Rate = distance/time
The function s gives the position of an object
as a function of time
Average velocity = change in distance
change in time
Average velocity = s / t
s = s(t +t) – s(t)
Find average velocity
of a falling object
If a billiard ball is dropped from a
height of 100 feet, its height s at time t
is given by the position function
s = -16t2 + 100
s(t) is the position function of the
billiard ball measure in feet
t = time measured in seconds
100 = “ORIGIN”AL HEIGHT
aka Initial Height
Find the average velocity
s(t) = -16t2 + 100
find average velocity over the time
period [1,2]
s(1) = 84 feet and s(2) = 36 feet
So average velocity is –48 ft/s
s 36  84             Why is it 36 – 84 ?
        48
t   2 1              Why is the velocity
negative?
Velocity                        s (t  t )  s (t )
v(t )  lim
function                   t 0          t
LOOKS LIKE THE DERIVATIVE!!!!! 

So the velocity function is the Derivative of the position
function !!!!! 

YEAH!!!
Average velocity vs
instant velocity
Average velocity between t1 and t2 is
the slope of the secant line
Instantaneous velocity at t1 is the slope
of your tangent line
Position function of a FREE
falling object s(t )  1 gt 2  v0t  s0
2
Neglecting air resistance….
s0 = initial height of the object
v0 = initial velocity of the object
g~ -32 ft/s2 or –9.8 m/s2 (acceleration
due to gravity on earth)
Example
At time t=0, a diver jumps from a platform
diving board that is 32 feet above water. The
position of the diver is given by the following
position function:
s(t )  16t  16t  32
2

Where s is measured in feet and t is
measured in seconds.
When does the diver hit the water?
What is the diver’s velocity at impact?
Example cont’d s(t )  16t  16t  32
2

32 is the initial height (height of board
above water)
From the middle term, 16t, 16 is the
initial velocity of the diver
To find the time t when the diver hits the
water, let s = 0 and solve for t. If s = 0
then the position is 0, right b/c the
diver HITS the water…..
Example cont’d s(t )  16t  16t  32
2

So we let s = 0 and solve for t to find
the time it takes for the diver to hit the
water                    t can not be negative…
no negative time… this
0  16t  16t  32
2
is not back to the
future, ok? 
0  16(t  t  2)
2
So t = 2 is the only
0  16(t  2)(t  1)
At t = 2 seconds, the
t  1, t  2              diver hits the water –
that’s fast!
Example cont’d s(t )  16t  16t  32
2

Next, lets solve for the diver’s velocity at
impact.
We use t=2, b/c we just found out that’s the
time it takes for the diver to hit the water and
we want velocity at impact (like you know
when the diver hits the water, duh)
Remember, velocity is the derivate of position
s '(t )  32t  16
s '(2)  64  16  48 ft / s

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