Handout 4, Summer 2014
Math 1823-171
28 May 2014
d
(4x − x2 )100
dx
Solution. Let
1.
f (x) = x100
g(x) = 4x − x2
Then (4x − x2 )100 = (f ◦ g)(x). Further, we have
f 0 (x) = 100x99
g 0 (x) = 4 − 2x
Therefore, we have
d
d
(4x − x2 )100 =
(f ◦ g)(x)
dx
dx
= f 0 (g(x)) · g 0 (x)
= 100(4x − x2 )99 (4 − 2x)
This gives the solution.
d√
1 − 2x
dx
Solution. Let
2.
f (x) =
√
x = x1/2
g(x) = 1 − 2x
Then
√
1 − 2x = (f ◦ g)(x). Further, we have
1
f 0 (x) = x−1/2
2
0
g (x) = −2
Therefore, we have
d√
d
1 − 2x =
(f ◦ g)(x)
dx
dx
= f 0 (g(x)) · g 0 (x)
1
−1/2
(1 − 2x)
= −(1 − 2x)−1/2
= −2
2
This gives the solution.
1
2
d
3.
dx
x2 + 1
x2 − 1
5
Solution. Let
f (x) = x5
g(x) =
Then
x2 +1
x2 −1
5
x2 + 1
x2 − 1
= (f ◦ g)(x). Further, by the quotient rule we have
f 0 (x) = 5x4
g 0 (x) =
(2x)(x2 − 1) − (x2 + 1)(2x)
(x2 − 1)2
Therefore, we have
5
d x2 + 1
d
=
(f ◦ g)(x)
2
dx x − 1
dx
= f 0 (g(x)) · g 0 (x)
2
4
x +1
(2x)(x2 − 1) − (x2 + 1)(2x)
=5
x2 − 1
(x2 − 1)2
This gives the solution.
d 2
(x + 1)3 (x2 + 2)6
dx
Solution. Note first that we have a product of two functions. Therefore,
the product rule gives
d 2
d
d
(x + 1)3 (x2 + 2)6 = (x2 + 2)6 (x2 + 1)3 + (x2 + 1)3 (x2 + 2)6
dx
dx
dx
Using the chain rule, we have
d 2
(x + 1)3 = 3(x2 + 1)2 (2x) = 6x(x2 + 1)2
dx
and
d 2
(x + 2)6 = 6(x2 + 2)5 (2x) = 12x(x2 + 2)5
dx
Therefore, we have
d 2
(x + 1)3 (x2 + 2)6 = (x2 + 2)6 6x(x2 + 1)2 + (x2 + 1)3 12x(x2 + 2)5
dx
4.
3
5.
d
x
√
dx 7 − 3x
Solution. This time, we have a quotient of two functions. By the quotient rule, we have
√
√
d
d
x − x dx
7 − 3x dx
7 − 3x
d
x
√
√
=
2
dx
7 − 3x
( 7 − 3x)
√
d
d
Now, dx
x = 1, but to get dx
7 − 3x, we must use the chain rule to get
d√
1
−3
7 − 3x = (7 − 3x)−1/2 (−3) =
(7 − 3x)−1/2
dx
2
2
Therefore, overall we have
√
−1/2
7 − 3x − x −3
(7
−
3x)
d
x
√
√ 2
=
dx
7 − 3x
( 7 − 3x)2
d
sin2 (2x)
dx
Solution. One can complete this using the product rule, but we will
instead use the chain rule twice. For the first chain, we have
6.
f1 (x) = x2
g1 (x) = sin(2x)
f10 (x) = 2x
To find g10 (x), we must again use the chain rule. Therefore, we have
f2 (x) = sin(x)
g2 (x) = 2x
f20 (x) = cos(x)
g20 (x) = 2
Therefore, by the chain rule, we have that g10 (x) = 2 cos(2x). Plugging
this back into the first chain rule, we arrive at:
d
sin2 (2x) = 2(sin(2x))2(cos(2x)) = 4 sin(2x) cos(2x)
dx
7.
3
d
x + sin2 (x)
dx
4
Solution. Again, we solve this by using the chain rule twice. For the
first chain, we have
f1 (x) = x3
g1 (x) = x + sin2 (x)
f10 (x) = 3x2
Once again, to find g10 (x), we need to use the chain rule. We have
f2 (x) = x2
g2 (x) = sin(x)
f20 (x) = 2x
g20 (x) = cos(x)
Therefore, we have that (sin2 (x))0 = 2 sin(x) cos(x), so that g10 (x) =
1 + 2 sin(x) cos(x). Applying this to the first chain rule, we have
3
d
x + sin2 (x) = 3(x + sin2 (x))2 (1 + 2 sin(x) cos(x))
dx
We have the result.
8. Suppose that h(x) =
4. Find h0 (1).
p
4 + 3f (x), where f (1) = 7 and f 0 (1) =
Solution. Using the chain rule, we first have that
1
3f 0 (x)
h0 (x) = (4 + 3f (x))−1/2 (3f 0 (x)) = p
2
2 4 + 3f (x)
(This is where the ability to simplify algebraic expressions is useful).
Plugging in, we have
3f 0 (1)
3·4
6
h0 (1) = p
= √
=
5
2 4+3·7
2 4 + 3f (1)
9. Find all points on the graph of f (x) = 2 sin(x) + sin2 (x) at
which the tangent line is horizontal.
Solution. Recall that the derivative at a point is the slope of the tangent
line at that point. Furthermore, a tangent line is horizontal if and only
if its slope is 0. Therefore, the points at which the tangent line to the
graph is horizontal are precisely the points where the derivative is 0.
5
Therefore, we first calculate the derivative. Using the chain rule on the
second term of f (x), we have
f 0 (x) = 2 cos(x) + 2 sin(x) cos(x) = 2 cos(x)(1 + sin(x))
Setting this derivative equal to 0, we see that this is the case precisely
when cos(x) = 0 or sin(x) = −1. The former occurs for x = π2 + 2nπ
+ 2nπ for any integer n. The latter occurs
for any integer n or x = 3π
2
3π
for x = 2 + 2πn for any integer n. Therefore, the tangent line to the
graph is horizontal at each of these x-values.
What are the corresponding points on the graph? Well, if x = π2 +
+ 2nπ for
2nπ for some integer n, then f (x) = 2 + 1 = 3. If x = 3π
2
some integer n, then f (x) = −2 + 1 = −1. Therefore, the points on
the graph at which the tangent line is horizontal are all points of the
form:
π
3π
+ 2nπ, 3
+ 2nπ, −1
2
2
This gives the result.