# Math 1210-1 Midterm 1

```Math 1210-1 Midterm 1
Solution
1. (10 points) Find the following derivatives. You don’t have to show your work.
(a) Dx (x1025 ) = 1025x1024
√
1 3
1
(b) Dx ( 4 x) = Dx (x 4 ) = x− 4
4
(c) Dx (x−3 ) = −3x−4
(d) Dx (2x3 + 3x2 + 5x − 1) = 6x2 + 6x + 5
(e) Dx401 (sin x) = Dx4&times;100+1 (sin x) = Dx1 (sin x) = cos x
2. (10 points) Suppose

2

if x &lt; 0
 x
f (x) =
1
if x = 0


x − 1 if x &gt; 0
Find the following value and limits. You don’t have to show your work.
(a) f (0) = 1
(b) lim f (x) = -1
x→0+
(c) lim− f (x) = 0
x→0
(d) lim f (x) = DNE
x→0
(e) lim f (x) = 0
x→1
3. (10 points) Find the following limit:
cos x
x→0 x2
lim
Solution: Since cos 0 = 1 and 02 = 0, this is a limit of the type nonzero
. So the limit must be ∞
0
or −∞. Suppose x is close to 0 but not 0. Then cos x is close to 1, hence positive. On the other
hand, x2 is always positive when x 6= 0. So the fraction is always positive. The limit is ∞ .
4. (10 points) Find the following limit:
x2 + 2x − 3
x→1
x2 − 1
lim
Solution:
x2 + 2x − 3
(x − 1)(x + 3)
= lim
x→1
x→1 (x − 1)(x + 1)
x2 − 1
x+3
= lim
x→1 x + 1
1+3
=
1+1
lim
= 2
1
5. (10 points) Find the following derivative:
Dx
Solution:
Dx
x2
x−1
x2
x−1
=
Dx (x2 ) &middot; (x − 1) − Dx (x − 1) &middot; x2
(x − 1)2
=
2x(x − 1) − x2
(x − 1)2
=
x2 − 2x
(x − 1)2
6. (10 points) Find the following limit:
x4 + x2 + 1
x→∞ x3 − x + 100
lim
Solution:
4
2
x +x +1
= lim
x→∞
x→∞ x3 − x + 100
lim
=
=
=
x4 +x2 +1
x4
&middot;
x4
x3 −x+100
x3
x3
1 + x12 + x14
lim
&middot;x
x→∞ 1 − 12 + 100
x
x3
2
4
1 + x1 + x1
lim
x→∞ 1 − 1 2 + 100 1 3
x
x
&middot;x
1+0+0
&middot;∞
1+0−0
=1&middot;∞
= ∞
7. (a) (5 points) Find dy/dx if y is given by the following equation x2 + y 2 = 1, i.e., the unit circle.
Solution: Take Dx on both sides of this equation.
2x + 2yy 0 = 0
2yy 0 = −2x
y0 =
−2x
2y
y0 = −
x
y
(b) (3 points) Find the slope of tangent line of the unit circle at
√
3 1
2 , 2
.
√
3
and
Solution: Since the slope of tangent line is the value of the derivative, we plug in x =
2
1
x
y = to − that we found in part (a). We conclude that the slope of tangent line is
2
y
√ 3
√
2
− 1 = − 3
2
2
(c) (2 points) Find the slope of the radius from the origin to
between this radius and the tangent line in (b)?
√
Solution: The slope of the line through (0, 0) and 23 , 12 is
y1 − y 2
=
x1 − x2
1
−0
√2
3
2 −0
√
3 1
2 , 2
. What is the relation
1
= √
3
Since this slope is the negative reciprocal of the slope of tangent line that we found in (b),
these two lines are perpendicular to each other.
8. (10 points) Find the tangent line of f (x) = sin(πx) at (1,0).
Solution: Suppose the equation of the tangent line is y = mx + b. Then the slope m is f 0 (x) at
x = 1. Since f 0 (x) = π cos(πx). We have m = f 0 (1) = π cos(π) = −π. On the other hand, the line
must pass the point (1,0). So 0 = −π &middot; 1 + b, or b = π. To conclude, the equation of the tangent
line is y = −πx + π .
9. (a) (5 points) Estimate
√
8.97
√
Solution: Suppose f (x) = x. We are going to estimate f (8.97). We know f (9) = 3 and
8.97 = 9 − 0.03. Apply the formula
f (x + ∆x) ≈ f (x) + f 0 (x) &middot; ∆x
1
where x = 9, ∆x = −0.03 and f 0 (x) = √ . We get
2 x
f (8.97) ≈ 3 +
(b) (5 points) Estimate
√
3
1
&middot; (−0.03) = 3 − 0.005 = 2.995
2&middot;3
8.03
√
Solution: Suppose f (x) = 3 x. We are going to estimate f (8.03). We know f (8) = 2 and
8.03 = 8 + 0.03. Apply the formula
f (x + ∆x) ≈ f (x) + f 0 (x) &middot; ∆x
1
where x = 8, ∆x = 0.03 and f 0 (x) = √
. We get
3
3 x2
f (8.03) ≈ 2 +
1
&middot; 0.03 = 2 + 0.0025 = 2.0025
3&middot;4
10. (Bonus) Suppose that f (x) is an odd function and the positive part of the graph is as follows:
b
a
-a
b
f(x)
3
(a) (2 points) Sketch the negative part of the graph.
Solution: This is shown as the black curve on the left side of the graph.
(b) (2 points) Sketch the tangent lines at (a, 0) and (−a, 0), respectively.
Solution: These are shown as the red lines on the graph.
(c) (3 points) Suppose that f 0 (a) = −1. Find f 0 (−a).
Hint: You are going to use the identity f (−x) = −f (x) for odd functions.
Solution: Since f (x) is an odd function, we have
f (−x) = −f (x).
Just as implicit differentiation, we take Dx on both sides of this equation and get
−f 0 (−x) = −f 0 (x),
where the left hand side is by the chain rule. We plug in x = a and obtain
−f 0 (−a) = −f 0 (a) = −(−1) = 1.
Thus,
f 0 (−a) = −1
This is obvious from the graph since the two tangent lines are parallel to each other.
(d) (3 points) Is f 0 (x) an even/odd/neither function? Circle your answer and justify it.
Hint: Compare f 0 (−x) with f 0 (x).
Solution: In part (c), we obtained the equation
−f 0 (−x) = −f 0 (x).
This is equivalent to
f 0 (−x) = f 0 (x).
Therefore f 0 (x) is an even function. In general, the derivative of an odd function is even and
the derivative of an even function is odd.
4
```