Calculus 1, Fall 2014
Solutions to Homework 5
Maximal grade for HW5: 40 points
Section 3.4: 34. (10 points) Find the derivative of the function f (x) =
2 −1/x
xe
.
Solution: By the Product Rule and the Chain Rule, one has
f 0 (x) = (x2 )0 e−1/x + x2 (e−1/x )0 = 2xe−1/x + x2 e−1/x (−1/x)0 =
2xe−1/x + x2 e−1/x (1/x2 ) = 2xe−1/x + e−1/x = (2x + 1)e−1/x .
Section 3.5: 14. (10 points) Find y 0 (x) by implicit differentiation from
the equation
ey sin x = x + xy.
Solution: Let us take the derivatives of both sides of the equation:
ey · y 0 sin x + ey cos x = 1 + y + xy 0 ,
and we can solve the resulting linear equation for y 0 :
y 0 (ey sin x − x) = 1 − ey cos x + y, y 0 =
1 − ey cos x + y
.
ey sin x − x
34. (10 points) For the curve with equation y 2 = x3 + 3x2 :
a) Find an equation of the tangent line at point (1, −2)
b) At what points does this curve have horizontal tangents?
c) Sketch the curve and the tangent lines from (a) and (b).
1
Solution: a) Let us take the derivatives of the both sides of the equation:
2yy 0 = 3x2 + 6x, y 0 =
3x2 + 6x
.
2y
If x = 1 and y = −2 then y 0 = − 49 , so the equation of the tangent at
(1, −2) has a form y = − 94 x + b. Since it should pass through (1, −2), we
have −2 = − 49 + b, so b = 41 , and the equation of the tangent has the form
y = − 94 x + 14 .
b) We have y 0 = 0 when 3x2 + 6x = 0, that is, x = 0 or x = −2. Note
that at x = 0 we have y = 0, and y 0 is undefined. For x = −2 we have
y 2 = −8 + 12 = 4, so the
are y = ±2.
√ equations of two horizontal tangents
3
2
3
2
c) Note that y = ± x + 3x . The function f (x) = x + 3x = x2 (x − 3)
vanishes for x = 0 and x = −3, it is positive on (−3, 0) and on (0, +∞)
and negative
on (−∞, −3). Its graph is shown in Figure ??. The function
√
3
y = x + 3x2 is therefore defined for x ≥ −3, and its graph is shown in
Figure ??.
4
2
−4
−3
−2
−1
1
2
−2
−4
Figure 1: The graph y = x3 + 3x2
Section 3.6: 24. (10 points) Find y 0 and y 00 if y =
ln x
.
x2
Solution: We have
y0 =
(ln x)0 x2 − ln x(x2 )0
(1/x)x2 − ln x(2x)
x − 2x ln x
1 − 2 ln x
=
=
=
,
4
4
4
x
x
x
x3
2
4
2
−4
−3
−2
−1
1
2
−2
−4
Figure 2: The graph for problem 34
y 00 =
(1 − 2 ln x)0 x3 − (1 − 2 ln x)(x3 )0
(−2/x)x3 − (1 − 2 ln x)(3x2 )
=
=
x6
x6
6 ln x − 5
−2x2 − 3x2 + 6x2 ln x
=
.
6
x
x4
3