MATH 147
Lab Key #6
3/8/2016
(1) Differentiate each function.
(a) f ( x ) = cos (−5x ) − sin2 2x3 − 1 + cot 3x3 − 4x
sin (2t) + 1
(b) f (t) =
cos (6t) − 1
Solution:
(a)
i
i
d
d h 2 3
d h
sin 2x − 1 +
cot 3x3 − 4x
[cos (−5x )] −
dx
dx
dx
d h i
d h
i
d
= sin (−5x )
sin 2x3 − 1 − csc2 3x3 − 4x
3x3 − 4x
[−5x ] − 2 sin 2x3 − 1
dx
dx
dx
d h
i = −5 sin (−5x ) − 2 sin 2x3 − 1 cos 2x3 − 1
2x3 − 1 − 9x2 − 4 csc2 3x3 − 4x .
dx
= −5 sin (−5x ) − 12x sin 2x3 − 1 cos 2x3 − 1 − 9x2 − 4 csc2 3x3 − 4x .
f 0 (x) =
(b)
d
d
[sin (2t) + 1] · (cos (6t) − 1) − (sin (2t) + 1) · [cos (6t) − 1]
dt
dt
f (t) =
(cos (6t) − 1)2
0
cos (2t)
=
=
d
d
[2t] · (cos (6t) − 1) − (sin (2t) + 1) (− sin (6t)) [6t]
dt
dt
(cos (6t) − 1)2
2 cos (2t) [cos (6t) − 1] + 6 sin (6t) [sin (2t) + 1]
[cos (6t) − 1]2
.
1
(2) Differentiate each function.
(a) g ( x ) = e3√−5x cos ( x )
(b) f ( x ) = 3 2x+1
Solution: (a)
g0 ( x ) =
d h 3−5x i
d
e
cos ( x ) + e3−5x
[cos ( x )]
dx
dx
= e3−5x
d
[−5x ] cos ( x ) + e3−5x (− sin ( x ))
dx
= −5e3−5x cos ( x ) − e3−5x sin ( x ) .
(b)
i
d h√
2x + 1
dx
h
i
√
d
= (ln 3) · 3 2x+1 ·
(2x + 1)1/2
dx
√
1
d
= (ln 3) · 3 2x+1 · (2x + 1)−1/2 ·
[2x + 1]
2
dx
√
1
= (ln 3) · 3 2x+1 · (2x + 1)−1/2 · 2
2
f 0 ( x ) = (ln 3) · 3
√
2x +1
·
√
(ln 3) · 3 2x+1
√
=
.
2x + 1
(3) Differentiate each function.
(a) f ( x ) = ln 4x3 − 3x2 + 2x
(b) f ( x ) = log3 x4 − 3x
Solution: (a)
i
d h 3
4x − 3x2 + 2x
12x2 − 6x + 2
f 0 ( x ) = dx 3
= 3
.
2
4x − 3x + 2x
4x − 3x2 + 2x
(b)
i
d h 4
x − 3x
4x3 − 3
f 0 ( x ) = dx
=
.
(ln 3) ( x4 − 3x )
(ln 3) ( x4 − 3x )
2
(4) Let f ( x ) =
√
x2 + 1, x ≥ 0. Find f −1
0 √ 3 .
Solution: First, remember that
f −1
0
1
(x) =
f0
( f −1 ( x ))
.
Hence, to do this, we will find f 0 ( x ) . Since f ( x ) = x2 + 1
f 0 (x) =
1/2
,
−1/2 d h
i
1 2
x +1
x2 + 1 .
2
dx
f 0 (x) = √
x
x2 + 1
.
Now, we shall find f −1 ( x ) . To do this, we solve the following equation for y :
q
x = y2 + 1.
x2 = y2 + 1 =⇒ x2 − 1 = y2 =⇒ y = ±
p
x2 − 1.
However,
since the domain of our original function is non-negative, f −1 ( x ) =
√
2
x − 1. Therefore,
f −1
0 √ 3 =
f0
Now, f 0
√ 2 =
√
q √2
2
( 2 ) +1
f
−1
=
√
√2 .
3
r
1
1
! = √ .
√ 2
f0
2
3 −1
Therefore,
√
√
0 √ 0 √ 3
6
−
1
3 = √ or f
3 =
.
2
2
3
(5) (Radioactive Decay) Suppose W (t) denotes the amount of a radioactive
material left after t days. Assume that W (0) = 20 mg and dW
dt = −4W.
(a) Find an equation for the mass remaining after t days. Include the appropriate unit for mass.
(b) How long will it take the material to decay to 8 mg? Include the appropriate unit for time.
Solution: (a) We will use a model of the form W (t) = Pekt , where P is the
initial amount. Then
20 = W (0) = Pek·0 = P.
Therefore, W (t) = 20ekt . Now,
d h kt i
dW
=
20e .
dt
dt
d h kt i
e .
∴ −80ekt = 20
dt
d
∴ −80ekt = 20ekt [kt] .
dt
−4W =
∴ −80ekt = 20kekt .
80ekt
= k.
20ekt
Therefore, k = −4. Hence, the appropriate model for the described radioactive
decay is given by
W (t) = 20e−4t mg.
∴−
(b) Now we want to know what value of t makes W (t) = 8 mg. Therefore,
we need to solve this equation:
8 = 20e−4t .
2
= e−4t .
5
2
∴ ln
= −4t.
5
2
1
days.
∴ t = − ln
4
5
∴
4