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1. Find f 0 (x) where f (x) = 3x .
Solution: If we take ln in both side of this equation and differentiate it, we
get
ln f (x) = ln 3x
ln f (x) = x ln 3
d
d
(ln f (x)) =
(x ln 3)
dx
dx
1 0
f (x) = ln 3
f (x)
f 0 (x) = f (x) ln 3
f 0 (x) = 3x ln 3
∴
∴
∴
∴
∴
2. Find f 0 (x) where f (x) = (2x)1/x .
Solution: Similarly to the previous problem, we have
1
ln f (x) = ln 2x
x
d
d 1
∴
(ln f (x)) =
ln 2x
dx
dx x
1 0
−1
1 2
∴
f (x) = 2 ln 2x +
f (x)
x
x 2x
− ln(2x) + 1
0
∴ f (x) = f (x)
x2
− ln(2x) + 1
0
1/x
∴ f (x) = (2x)
x2
x
3. Find f 0 (x) where f (x) = (x)e .
Solution: We have
∴
∴
∴
ln f (x) = ex ln x
d x
d
(ln f (x)) =
(e ln x)
dx
dx
1 0
1
f (x) = ex ln x + ex
f (x)
x
0
ex
x
x1
f (x) = (x)
e ln x + e
x
4. Find f 0 (x) where f (x) = (x + 1)1−x .
Solution: We have
∴
∴
∴
ln f (x) = (1 − x) ln(x + 1)
d
d
(ln f (x)) =
((1 − x) ln(x + 1))
dx
dx
1 0
1
f (x) = − ln(x + 1) + (1 − x)
f (x)
x+1
1−x
0
1−x
f (x) = (x + 1)
− ln(x + 1) +
x+1
1
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´
5. Evaluate (2x + 7)5 dx.
Solution:
ˆ
(2x + 7)5 dx =
(2x + 7)6
(2x + 7)6
+C =
+C
6∗2
12
where C ∈ R.
´
6. Evaluate (ax + b)k dx for all k ∈ R (assume a 6= 0).
Solution: If k 6= 1 we have
ˆ
(ax + b)k+1
(ax + b)k dx =
+C
(k + 1)a
where C ∈ R.
For k = −1 we have
ˆ
(ax + b)−1 dx = ln |ax + b| + C
where C ∈ R.
7. Find f such that f 0 (x) = e5x .
e5x
+ C, where C ∈ R.
Solution: f (x) =
5
2
8. Find f such that f 0 (x) = 3/2 and f (1) = 5.
x
x−3/2+1
2
0
+ C = −4x−1/2 + C,
Solution: Since f (x) = 3/2 , we have f (x) = 2
x
−3/2 + 1
for some constant C ∈ R. Imposing f (1) = 5 we get
5 = −4(1)−1/2 + C
⇒
5 = −4 + C
⇒
C = 9.
Therefore, f (x) = −4x−1/2 + 9.
9. Find f such that f 00 (x) =
1
,
x2
f (2) = 0 and f (1) = 1.
Solution: Here we need to integrate twice.
Since f 00 (x) = x12 , then integrating the equation once we get f 0 (x) =
− x1 + C1 , for some constant C1 .
x−1
−1
+C1 =
Now since f 0 (x) = − x1 + C1 , integrating this equation we get f (x) = − ln |x| +
C1 x + C2 , for some constants C1 and C2 .
Using f (x) = − ln |x| + C1 x + C2 , and setting f (2) = 0 and f (1) = 1 we get
∴
∴
∴
(
− ln 2 + 2C1 + C2 = 0
C1 + C2 = 1
(
2C1 + C2 = ln 2
(I)
C1 + C2 = 1
(II)
C1 = ln 2 − 1
C2 = 2 − ln 2
2
from (I) − (II)
from (II)
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Therefore f (x) = − ln |x| + (ln 2 − 1)x + (2 − ln 2) (check that this f satisfy all
conditions f 00 (x) = x12 , f (2) = 0 and f (1) = 1).
10. Find the area enclosed by the curves y = x2 − 4 and y = 4 − x2 .
5
x2−4
4−x2
4
3
2
1
0
−1
−2
−3
−4
−5
−3
−2
−1
0
1
2
3
By symmetry, we note that the area A enclosed by the curves y = x2 − 4 and
y = 4 − x2 is
ˆ
2
8
64
8
x3 x=2
= 2 8 − − (−8 + ) =
A=2
4 − x dx = 2 4x −
3
3
3
3
x=−2
−2
2
You could also have written A =
´2
−2
4 − x2 dx −
´2
−2
x2 − 4dx.
11. (in the worksheet given in class the function was x2 − 4x + 5) Find the area
under the curve y(x) = x2 − 6x + 5 from x = −1 to x = 3.
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12
f(x)=x2−6x+5
10
8
6
4
2
0
−2
−4
−1
0
1
2
3
4
5
6
We note that the area is given by
ˆ
ˆ
1
3
(x − 6x + 5)dx −
(x2 − 6x + 5)dx =
−1
1
x=1
x=3
3
2
3
2
= (x /3 − 3x + 5x)
− (x /3 − 3x + 5x)
= 16
A =
2
x=−1
x=1
12. (Textbook, p. 289, ex. 13) The population of a certain country is growing
exponentially. The total population (in millions) in t years is given by the
function P (t). Match each of the following answers with its corresponding
question.
Answers:
(a) Solve P (t) = 2 for t.
(e) y 0 = ky
(b) P (2)
(f) Solve P (t) = 2P (0) for t
(c) P 0 (2)
(g) P0 ekt , k > 0
(d) Solve P 0 (t) = 2 for t
(h) P(0)
Questions:
(I) How fast will be the population growing in 2 years?
(II) Give the general formula of the function P (t).
(III) How long will it take for the current population to double?
(IV) What will be the size of the population in 2 years?
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(V) What is the initial size of the population?
(VI) When will be the size of the population 2 million?
(VII) When will be the population growing at the rate of 2 million people per
year?
(VIII) Give the differential equation satisfied by P (t).
Solution: (a)-(VI); (b)-(IV); (c)-(I); (d)-(VII); (e)-(VIII); (f)-(III); (g)-(II);
(h)-(V)
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