1.6 Topics: Using basic differentiation rules. SOLUTIONS
Use the rules of differentiation to find the derivatives. Hint: Rewrite #5 – 7 before differentiating.
1. f ( x ) = 5
f ′( x) = 0
6. f ( x ) =
2
g ′ (θ ) =
y′ = 6 x − 5
5. y = 4 x
π
3. g (θ ) =
2. y = 3x 2 − 5 x + 1
3
4 x2
7. g ( x ) =
π
2
4. h ( t ) = 2et − 3cos t
sin θ − cos θ
h′ ( t ) = 2e t + 3sin t
cos θ + sin θ
4
x3 − 5 x 2 + 4 x − 2
x2
3
8. h ( x ) = x 5 − x 5
x3 − 5 x 2 + 4 x − 2
x2
3
x 5x2 4 x 2
g ( x) = 2 − 2 + 2 − 2
x
x
x
x
−1
−2
g ( x ) = x − 5 + 4x − 2x
g ( x) =
y=x
y′ =
1
4
1 − 34
x =
4
3 x −2
4
3
3
f ′ ( x ) = ⋅ ( −2 ) x −3 = − 3
4
2x
f ( x) =
1
4x
3
4
g ′ ( x ) = 1 − 0 + 4 ( −1) x −2 − 2 ( −2 ) x −3
g′( x) = 1−
4 − 15 3 − 25
x − x
5
5
2
1
−


1
h′ ( x ) = x 5  4 x 5 − 3 
5


h′ ( x ) =
4
4
+
x 2 x3
h′ ( x ) =
3
g′( x) =
x − 4x + 4
x3
45 x − 3
55 x2
9. What is the equation of the line tangent to f ( x ) = x 2 − 3 x − 4 at the point (1, −6 ) ?
f ( x ) = x 2 − 3x − 4
f ′ ( x ) = 2 x − 3 so the slope of the tangent line at (1, −6 ) will be m = 2 (1) − 3 = −1
So the equation of the tangent line will be....
y = −1x + b → − 6 = −1(1) + b → b = −5
y = −x − 5
10. At what point(s)
( x, y )
will y = 3 x + 2 cos x , 0 ≤ x < 2π , have a horizontal tangent line?
To solve this problem we will take the derivative and set it equal to zero (a horizontal tangent line will have a slope equal to zero).
y = 3x + 2 cos x , 0 ≤ x < 2π ,
y′= 3 − 2sin x → 0 = 3 − 2sin x → sin x =
3
2
→ x=
π 2π
,
3 3
 π 3π

So the "points" where there will be a horizontal tangent line are...  ,
+ 1 and
3 3

 2π 2 3π

,
− 1

3
 3

Note: To get the “y” values for the points just substitute the “x” values into the original equation y = 3 x + 2 cos x
11. If f ( x ) = g ( x ) + 3 what is the relationship between the derivative of f ( x ) and the derivative of g ( x ) ?
If two functions differ by only a constant then their derivatives will be equal ( f ( x ) = g ( x ) + 3 means that the graph of f ( x ) will be
exactly three units higher than the graph of g ( x ) at every value of x in the domain. So for a particular value of x the slope of the
tangent line to the graph of f ( x ) will be exactly the same as the slope of the tangent line to the graph of g ( x ) at that same x value).
f ( x) = g ( x) + 3 →
d
d
f ( x ) ) = ( g ( x ) + 3) →
(
dx
dx
f ′( x) = g′( x) +
d
( 3) →
dx
f ′( x) = g′( x) + 0 →
f ′( x) = g′( x)
12. If f ( x ) = −3 ⋅ g ( x ) what is the relationship between the derivative of f ( x ) and the derivative of g ( x ) ?
Due to the constant multiple rule we know that the derivative of f ( x ) will be −3 times the derivative of g ( x ) .
f ( x ) = −3 ⋅ g ( x ) →
d
d
( f ( x ) ) = dx ( −3 ⋅ g ( x )) →
dx
f ′ ( x ) = −3 ⋅ g ′ ( x )
From a graphical perspective, if f ( x ) = −3 ⋅ g ( x ) then the graph of f ( x ) is a reflection of the graph of g ( x ) about the x – axis and
stretched vertically by a factor of three. So for any particular x value, the slope of the tangent line at x on the graph of f ( x )
will be – 3 times what it was at the same x value on the graph of g ( x ) .
For a free falling object s ( t ) = −16t 2 + v0 t + s0 gives the position (in feet) of an object with initial velocity v0 (in feet per second) and
initial position s0 where t is time (in seconds).
A ball is thrown straight down from the top of a 220 foot building with an initial speed of 22 feet per second.
(Any answers that are not exact please round to two decimal places).
13. What is the velocity of the ball after three seconds?
s ( t ) = −16t 2 + v0 t + s0
→ s ( t ) = −16t 2 − 22t + 220
ft
s
v ( t ) = s′ ( t ) = −32t − 22 → v ( 3) = −32 ( 3) − 22 = −118
14. What is it’s velocity after falling 108 feet?
To solve this just set the position (s(t)) equal to 108 ft to find out WHEN (what time) this occurs then evaluate v(t) at this value of
time.
s ( t ) = −16t 2 − 22t + 220 → 108 = −16t 2 − 22t + 220 → 16t 2 + 22t − 112 = 0 → 2 ( 8t 2 + 11t − 56 ) = 0
2
8t + 11t − 56 = 0 → t =
t=
(11)
−11 ±
2
− 4 ( 8 )( −56 )
2 (8)
=
−11 ± 121 + 1792 −11 ± 1913
=
16
16
−11 + 1913
≈ 2.05 seconds
16
v ( 2.05 ) = −32 ( 2.05 ) − 22 = −87.6
ft
s
15. What is the speed of the ball when it hits the ground?
To solve this set the position function equal to zero to find out WHEN this happens. Then evaluate the velocity function at this time
AND then take the absolute value of velocity to get speed.
s ( t ) = −16t 2 − 22t + 220 → 0 = −16t 2 − 22t + 220 → 16t 2 + 22t − 220 = 0 → 8t 2 + 11t − 110
t=
−11 ±
(11)
2
− 4 ( 8 )( −110 )
2 (8)
=
−11 ± 121 + 3520 −11 ± 3641
−11 + 3641
=
→ t=
≈ 3.08 seconds
16
16
16
Speed ( 3.08 ) = v ( 3.08 ) = −32 ( 3.08 ) − 22 = −120.56 = 120.56
ft
s
For a free falling object s ( t ) = −4.9t 2 + v0 t + s0 gives the position (in meters) of an object with initial velocity v0 (in meters per
second) and initial position s0 where t is time (in seconds).
A projectile is shot upward from the surface of the earth with an initial velocity of 120 meters/second.
(Any answers that are not exact please round to two decimal places).
16. What is the velocity of the projectile after 5 seconds?
s ( t ) = −4.9t 2 + v0 t + s0
→ s ( t ) = −4.9t 2 + 120t + 0 → s ( t ) = −4.9t 2 + 120t
v ( t ) = s ′ ( t ) = −9.8t + 120 → v ( 5) = −9.8 ( 5 ) + 120 = 71
m
s
17. How high does the projectile go?
At the peak the projectile will temporarily have a velocity of zero (before it starts heading back down towards the surface of the earth).
So just set the velocity function equal to zero and solve for time. Next evaluate the position function at this time to find the max
height.
v ( t ) = −9.8t + 120 → 0 = −9.8t + 120 → t =
120
≈ 12.24 seconds
9.8
2
s (12.24 ) = −4.9 (12.24 ) + 120 (12.24 ) = 734.69 m
18. Find the equations of the tangent lines to the curve y = x3 − 9 x through the point (1, −9 )
y = x3 − 9 x →
y ′ = 3x 2 − 9 . Let ( a, b ) be a point of tangency on the graph of y = x 3 − 9 x described in the problem.
We know that the slope at ( a, b ) will be m = 3a 2 − 9. If the line is also going to pass through the point (1, −9 ) then the slope
can also be expressed using the slope formula... m =
b − −9 b + 9
. Setting the two slope expressions equal to one another
=
a −1
a −1
b+9
→ ( 3a 2 − 9 ) ( a − 1) = b + 9 → 3a 3 − 3a 2 − 9a + 9 = b + 9. In addition since the point ( a, b )
a −1
lies on the graph of y = x3 − 9 x we know that b = a 3 − 9a so substituting into the previous equation we get the following...
yeilds 3a 2 − 9 =
3
3a 3 − 3a 2 − 9a + 9 = a 3 − 9a + 9 → 2a 3 − 3a 2 = 0 → a 2 ( 2a − 3) = 0 → a = 0, a = . Using the function y = x3 − 9 x
2
3
 3 81 
we find that the y values that are partnered with x = 0 and x = and obtain the points ( 0, 0 ) and  , −  and the slope
2
2 8 
2
2
of the tangent line at these points will be m = 3 ( 0 ) − 9 = −9 and
So the equations of the two tangent lines will be
y = −9 x
and
9
27
y = − x−
4
4
9
3
m = 3   − 9 = − respectively.
2
4
 