Transcript Document

2.2 Differentiation Rules for Constant Multiples,
Sums, Powers, Sines, and Cosines
Constant Rule:
The derivative of a constant is zero.
d
[c ]  0
dx
Find the derivatives of:
y'  0
y7
f ( x)  0
f ' ( x)  0
s' (t )  5
s" (t )  0
Power Rule: If n is a rational number, then
d n
n 1
[ x ]  nx
dx
Find the derivatives of:
f ( x)  x
f ' ( x)  3x 2
3
1
y  2 rewritten as
x
g ( x)  x
x
2
y'  2 x
g ' ( x)  1
3
2
 3
x
s(t )  16t  64t  100 s' (t )  v(t )  32t  64
2
Differentiate:
dy
2
2
2
1
 2 x   2
y
 2x
x
dx
x
4t 2
8t
f (t ) 
f ' (t ) 
5
5
Sum and Difference Rules
3
d  3x 
d



dx  2 
2 dx  f ( x)  g ( x)  f ' ( x)  g ' ( x)
d
d
3x   3
 f ( x)  g ( x)  f ' ( x)  g ' ( x)
dx
dx
x4
3
g ( x)    3x  2 x
2
g ' ( x)  2x  9x  2
3
2
Differentiate:
y  2 x  2x
y
1
3
2 x
2
1
2
1 2 3
 x
2
1
1 12
1
2
y'  2  x  x 
2
x
1
dy 1  2  5 3
 5
   x
dx 2  3 
3x 3
Derivatives of Sine and Cosine
d
sin x  cos x
dx
y  3 sin x
y  x  cos x
d
cos x   sin x
dx
y ' 3 cos x
y'  1  sin x
Find the slope and equation of the tangent line
of the graph of y = 2 cos x at the point   ,1.


3 
f’(x) = -2sin x
 3

 
 3
@ f '    2 sin   2

3
2
3


Therefore, the equation of the tangent line is:


y  1   3 x  
3

Day 1
The average rate of change in distance with
respect to time is given by…
change in distance  s
change in time
t
Also known as
average velocity
Ex. If a free-falling object is dropped from a
height of 100 feet, its height s at time t is given
by the position function s = -16t2 + 100, where
s is measured in feet and t is measured in seconds.
Find the average rate of change of the height over
the following intervals.
a. [1, 2] b. [1, 1.5] c. [1, 1.1]
a.
b.
c.
s 36  84  48

 48 ft / sec

2 1
1
t
s 64  84  20
 40 ft / sec


1.5  1
.5
t
s 80 .64  84  3.36
 33.6 ft / sec


.1
1 .1  1
t
At time t = 0, a diver jumps from a diving board
that is 32 feet above the water. The position of the
diver is given by
s(t )  16t  16t  32
2
where s is measured in feet and t in seconds.
a. When does the diver hit the water?
b. What is the diver’s velocity at impact?
To find the time at which the diver hits the water,
we let s(t) = 0 and solve for t.
0  16t  16t  32
0  16t  1t  2
0  16 t 2  t  2
t = -1 or 2
2


-1 doesn’t make sense, so the diver hits at 2 seconds.
The velocity at time t is given by the derivative.
s’(t) = v(t) = -32t + 16
@ t = 2 seconds,
s’(2) = -48 ft/sec.
The negative gives the direction, which in this
case is down.