Math 113 HW #6 Solutions
1. Exercise 3.1.22. Differentiate the function
y=
√
x(x − 1)
Answer: Re-write the function as
√
√
y = x x − x = x3/2 − x1/2 .
Then, using the power rule,
3
1
1
3√
y 0 = x1/2 − x−1/2 =
x− √ .
2
2
2
2 x
2. Exercise 3.1.34. Find an equation of the tangent line to the curve
y = x4 + 2x2 − x
at the point (1, 2).
Answer: Using the power rule,
y 0 = 4x3 + 4x − 1,
so, at the point (1, 2),
y 0 = 4(1)3 + 4(1) − 1 = 7.
Therefore the slope of the tangent line is 7. Since the tangent line passes through the point
(1, 2), we can use the point-slope formula:
y − 2 = 7(x − 1),
so, adding 2 to both sides, we have that
y = 7x − 5.
3. Exercise 3.2.24. Differentiate
f (x) =
1 − xex
.
x + ex
Answer: Using the quotient and product rules,
(x + ex )(−ex − xex ) − (1 − xex )(1 + ex )
(x + ex )2
−xex − x2 ex − e2x − xe2x − (1 + ex − xex − xe2x )
=
(x + ex )2
2
x
x
−1 − (1 + x + e )e
=
.
(x + ex )2
f 0 (x) =
1
4. Exercise 3.2.28. Let
f (x) = x5/2 ex
and find f 0 (x) and f 00 (x).
Answer: Using the product rule,
5 3/2 x
5
0
f (x) =
x
e + x5/2 ex = x3/2 ex + x5/2 ex .
2
2
(1)
Taking the derivative of the first term yields
15
5
5 3 1/2 x
x e + x3/2 ex = x1/2 ex + x3/2 ex ,
2 2
4
2
whereas the second term in (1) is just a copy of f , so its derivative is the same as f 0 (x).
Therefore,
15 1/2 x 5 3/2 x
00
x e + x e + f 0 (x)
f (x) =
4
2
15 1/2 x 5 3/2 x
5 3/2 x
=
x e + x e +
x e + x5/2 ex
4
2
2
15
= x1/2 ex + 5x3/2 ex + x5/2 ex .
4
5. Exercise 3.2.34. Find the equations of the tangent line and normal line to the curve
√
x
y=
x+1
at the point (4, 0.4).
√
Answer: Writing x as x1/2 and using the quotient rule,
(x + 1) 21 x−1/2 − x1/2 (1)
0
y =
(x + 1)2
=
=
=
1 1/2
2x
+ 12 x−1/2 − x1/2
(x + 1)2
1 x−1/2 − x1/2
2 (x + 1)2
√
1
1 √x − x
2 (x + 1)2
Therefore, we can plug in x = 4 to get the slope of the tangent line at (4, 0.4):
√
1
1 √4 − 4
1 12 − 2
3
=
=−
.
2
2 (4 + 1)
2 25
100
Therefore, we can use the point-slope formula to get the equation of the tangent line:
y−
2
3
=−
(x − 4),
5
100
2
so the tangent line is
y=−
3
13
x+ .
100
25
On the other hand, the slope of the normal line is the negative reciprocal of the slope of the
tangent line, so the slope of the normal line is 100
3 . Therefore, the point-slope formula tells
us that the normal line is
2
100
y− =
(x − 4),
5
3
or
1994
100
x−
y=
3
15
6. Exercise 3.3.10. Differentiate
y=
1 + sin x
x + cos x
Answer: Using the quotient rule and what we know about the derivatives of sines and
cosines,
(x + cos x) cos x − (1 + sin x)(1 − sin x)
(x + cos x)2
2
x cos x + cos x − (1 + sin x − sin x − sin2 x)
=
(x + cos x)2
x cos x + cos2 x − 1 + sin2 x
=
.
(x + cos x)2
y0 =
Using the fact that sin2 x + cos2 x = 1, this simplifies as
y0 =
x cos x
.
(x + cos x)2
7. Exercise 3.3.22. Find an equation of the tangent line to the curve
y = ex cos x
at the point (0, 1).
Answer: Using the product rule,
y 0 = ex cos x + ex (− sin x) = ex cos x − ex sin x.
Therefore, plugging in x = 0 gives the slope of the tangent line at (0, 1):
e0 cos 0 − e0 sin 0 = 1 − 0 = 1.
Therefore, by the point-slope formula, the equation of the tangent line is
y − 1 = 1(x − 0),
or
y = x + 1.
3
8. Exercise 3.3.34. Find the points on the curve y = (cos x)/(2 + sin x) at which the tangent is
horizontal.
Answer: The tangent being horizontal means that the slope of the tangent line is zero. In
other words, we’re looking for those points where the derivative of the function is zero. Using
the quotient rule, the derivative is
y0 =
(2 + sin x)(− sin x) − cos x(cos x)
−2 sin x − sin2 x − cos2 x
=
.
(2 + sin x)2
(2 + sin x)2
Using the fact that sin2 x + cos2 x = 1, this simplifies to
y0 =
−2 sin x − 1
,
(2 + sin x)2
which is zero if and only if the numerator is zero. Therefore, we want to find the values of x
for which 0 = −2 sin x − 1 or, equivalently,
1
sin x = − .
2
This equality holds when
x = ...,
−5π −π 7π 11π 19π 23π
(12n − 5)π (12n − 1)π
,
, ,
,
,
,...,
,
,...
6
6 6
6
6
6
6
6
9. Exercise 3.4.12. Find the derivative of
f (t) =
√
3
1 + tan t.
Answer: Re-writing as
f (t) = (1 + tan t)1/3 = g ◦ h(t),
where
g(u) = u1/3
and h(t) = 1 + tan t,
then the Chain Rule tells us that
f 0 (t) = g 0 (h(t)) · h0 (t).
Now, g 0 (u) = 31 u−2/3 and h0 (t) = sec2 t, so
1
1
sec2 t
f 0 (t) = g 0 (h(t)) · h0 (t) = (h(t))−2/3 · sec2 t = (1 + tan t) sec2 t =
.
3
3
3(1 + tan t)2/3
10. Exercise 3.4.22. Find the derivative of the function
y = e−5x cos 3x.
Answer: By the Chain Rule, the derivative of h1 (x) = e−5x = f1 circg1 (x) where f1 (u) = eu
and g1 (x) = −5x is
h01 (x) = f10 (g1 (x)) · g10 (x) = eg1 (x) (−5) = −5e−5x .
4
Also using the Chain Rule, the derivative of h2 (x) = cos 3x = f2 ◦ g2 (x) where f2 (u) = cos u
and g2 (x) = 3x is
h02 (x) = f20 (g2 (x)) · g20 (x) = − sin(g2 (x)) · (3) = −3 sin 3x.
Therefore, using these facts and the product rule,
y 0 = −5e−5x cos 3x + e−5x (−3 sin 3x) = −5e−5x cos 3x − 3e−5x sin 3x.
11. Exercise 3.4.34. Find the derivative of the function
1
y = x sin .
x
Answer: By the Chain Rule, the derivative of h(x) = sin x1 = f ◦ g(x) where f (u) = sin u
and g(x) = 1/x = x−1 is
h0 (x) = f 0 (g(x)) · g 0 (x) = cos(g(x)) · (−x−2 ) = −
Therefore, using the product rule,
y 0 = 1 · sin
so
y 0 = sin
5
cos 1
1
− x · 2x ,
x
x
1
1
1
− cos .
x x
x
cos x1
.
x2