Exam 2 Material: Requested Solutions
Contents
1 Finding Tangent Lines
2
2 Section 5.5
2.1 Problem 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Problem 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Problem 33 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
4
5
3 Section 5.5
3.1 Problem 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Problem 35 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3 Problem 37 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
6
7
8
4 Chapter 3 Review
4.1 Problem 49 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Problem 50 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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9
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1
1
Finding Tangent Lines
Recall that the tangent line to the graph of y = f (x) at the point (a, f (a)) is the line which:
i) goes through the point (a, f (a))
ii) has slope m = f 0 (a)
Simple Example: Find the equation of the tangent line to f (x) = x2 at (−1, 1).
Solution: Here, a = −1. Observe that f (−1) = (−1)2 = 1, so the point on the graph we’re
talking about is (a, f (a)) = (−1, f (−1)) = (−1, 1), which is the point stated in the problem.
The slope of the tangent line through this point is given by m = f 0 (−1). We first find f 0 (x),
then plug in x = −1:
f (x) = x2 =⇒ f 0 (x) = 2x =⇒ f 0 (−1) = 2(−1) = −2
Hence, the tangent line in question has slope m = −2. Consequently, its equation can be written
y = −2x + b
for some b. To find b, we plug in the point (x, y) = (−1, 1):
(1) = −2(−1) + b =⇒ 1 = 2 + b =⇒ b = −1
Hence, the tangent line has equation:
y = −2x − 1
2
2
2.1
Section 5.5
Problem 23
√
Find the derivative of f (x) = e
x
.
Solution: Here, f is ”e to a function”. Hence, we can use the rule:
(eu )0 = eu · u0
We have:
f 0 (x)
√
=
(e
√
= e
√
= e
√
= e
x 0
)
x
√
· ( x)0
x
· (x 2 )0
x
·
1
1 −1
·x 2
2
=
1 − 1 √x
·x 2 ·e
2
=
e x
√
2 x
√
3
2.2
Problem 25
Find the derivative of f (x) = (x − 1)e3x+2 .
Solution: Here, f is a product, so we start with the product rule. The (x − 1) term will be
easy. For the e3x+2 term, we use the eu rule:
(eu )0 = eu · u0
We have:
0
f (x)
=
0
3x+2
(x − 1) · e
=
0
0
3x+2
3x+2
(x − 1) · e
+ (x − 1) · e
=
1 · e3x+2 + (x − 1) · e3x+2 · 3x + 2
=
1 · e3x+2 + (x − 1) · e3x+2 · 3
=
e3x+2 + 3xe3x+2 − 3e3x+2
=
3xe3x+2 − 2e3x+2
4
0
2.3
Problem 33
Find an equation of the tangent line to the graph of y = e2x−3 at the point ( 32 , 1).
Solution: (First, take a quick look at the ”Finding Tangent Lines” section above.)
Here, a = 32 and the slope of this tangent line is m = f 0 ( 23 ). We find f 0 (x) first, then plug in
3
0
2 . To find f , use the ”e to a function” rule:
f 0 (x)
=⇒
m = f0
=
(e2x−3 )0
=
e2x−3 · (2x − 3)0
=
e2x−3 · (2)
=
2e2x−3
3
=
2
2e2(3/2)−3
=
2e3−3
=
2e0
=
2(1)
=
2
Hence, using y = mx + b, with m = 2, we have:
y = 2x + b
Plugging in the point (x, y) = ( 32 , 1) we get:
3
(1) = 2
+ b = 3 + b =⇒ b = −2
2
Hence, the tangent line has equation:
y = 2x − 2
5
3
3.1
Section 5.5
Problem 19
Find the derivative of f (u) = ln(u − 2)3 .
Solution: First note that, while we sometimes use u to denote an ”inside function”, here it is
simply the name of the variable. The other thing is that, it can look a bit ambiguous, but the power
is only on the (u − 2), i.e.,
f (u) = ln(u − 2)3 = ln (u − 2)3
Hence, f is ”ln of a function” and we can use our rule:
(ln w)0 =
1
· w0
w
(There is a shortcut here, but we haven’t covered the relevant material.)
To find w0 , we will use the ”function to a (constant) power rule”:
(v n )0 = n · v n−1 · v 0
We have:
0
f (u)
=
0
3
ln (u − 2)
=
0
1
3
·
(u
−
2)
(u − 2)3
=
1
2
0
·
3
·
(u
−
2)
·
(u
−
2)
(u − 2)3
=
1
2
·
3
·
(u
−
2)
·
1
(u − 2)3
=
3 · (u − 2)2
(u − 2)3
=
3
u−2
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3.2
Problem 35
Find the second derivative of f (x) = ln 2x.
Solution:
=⇒
=⇒
f (x)
=
f 0 (x)
=
f 00 (x)
ln 2x
ln 2x
0
=
0
1
· 2x
2x
=
1
·2
2x
=
1
x
=
x−1
=
x−1
0
=
(−1) · x−2
=
−
7
1
x2
3.3
Problem 37
Find the second derivative of f (x) = ln(x2 + 2).
Solution:
=⇒
f (x)
=
f 0 (x)
=
=
=⇒
00
f (x)
ln(x2 + 2)
ln(x2 + 2)
x2
0
1
· x2 + 2
+2
=
1
· 2x
x2 + 2
=
2x
x2 + 2
=
0
2x
x2 + 2
0
=
(2x)0 (x2 + 2) − 2x(x2 + 2)0
(x2 + 2)2
=
2(x2 + 2) − 2x(2x)
(x2 + 2)2
=
2x2 + 4 − 4x2
(x2 + 2)2
=
4 − 2x2
(x2 + 2)2
= −
2x2 − 4
(x2 + 2)2
8
4
Chapter 3 Review
4.1
Problem 49
Find an equation of the tangent line to the graph of y =
Solution: Here, our function is f (x) =
√
√
√
4 − x2 at the point (1, 3).
√
4 − x2 and the point in question is (1, 3), i.e., ”a = 1”.
Hence, the slope of the tangent line we’re interested in equals f 0 (1). To find this, first find f 0 (x),
then plug in x = 1. To find f 0 (x), it helps to rewrite the function first.
p
1
f (x) = 4 − x2 = (4 − x2 ) 2
We will need to use the generalized power rule version of the chain rule, i.e, the ”function to a
(constant) power” rule:
(un )0 = n · un−1 · u0
We have
f 0 (x)
=
1 0
(4 − x2 ) 2
=
0
1
1
· (4 − x2 ) 2 −1 · 4 − x2
2
=
1
1
· (4 − x2 )− 2 · (−2x)
2
1
= −x · (4 − x2 )− 2
= −√
x
4 − x2
Hence,
m = f 0 (1) = − p
(1)
4 − (1)2
1
= −√
3
Thus, using ”y = mx + b”, we have:
1
y = −√ x + b
3
√
and pluggin in (x, y) = (1, 3) we have:
√ √
√
√
1
1
3
3
1
3
1
4
√
√
( 3) = −
·√ +√ =√ +√ =√
(1) + b =⇒ b = 3 +
=
1
3
3
3
3
3
3
3
Hence
1
4
y = −√ x + √
3
3
9
4.2
Problem 50
Find an equation of the tangent line to the graph of y = x(x + 1)5 at the point (1, 32).
Solution: (Again, make sure you look over the ”Finding Tangent Lines” section above.)
The slope of this tangent line is m = f 0 (1).
f 0 (x)
0
=
x · (x + 1)5
=
0
0
x · (x + 1)5 + x · (x + 1)5
4
0
= 1 · (x + 1) + x · 5 · (x + 1) · (x + 1)
5
4
= (x + 1) + x · 5 · (x + 1) · 1
5
=⇒
f 0 (1)
=
(x + 1)5 + 5x(x + 1)4
=
((1) + 1)5 + 5(1)((1) + 1)4
=
25 + 5 · 24
=
32 + 5 · 16
=
32 + 80
=
112
Hence, plugging in m and (x1 , y1 ) all at once:
(32) = 112(1) + b =⇒ b = −80
Thus, the tangent line has equation:
y = 112x − 80
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