MATH 220: WORKSHEET FOR 9/30 SOLUTIONS
(1) Use implicit differentiation to find
(a) x3 + 2xy = 7
Applying
d
dx
dy
dx
for each of the following curves:
dy
to both sides, we get 3x2 + 2y + 2x dx
= 0. Then
dy
dx
=
−3x2 −2y
2x
(b) x2 sin(y 4 ) + exy = arctan(x2 + 3y)
dy
d
Applying dx
to both sides, we get 2x sin(y 4 )+x2 cos(y 4 )4y 3 dx
=
dy
2x
3
4
3 2
4
So dx = ( (x2 +3y)2 +1 − 2x sin(y ))/(4y x cos(y ) − (x2 +3y)2 +1 )
dy
1
(2x+3 dx
).
(x2 +3y)2 +1
(c) ln(xy) + 6cos(x−y) = log5 (y 2 )
Applying
d
dx
to both sides, we get
dy
dy
1
dy
1
(y + x ) + ln(6)6cos(x−y) (− sin(x − y))(1 − ) =
2y
2
xy
dx
dx
ln(5)y
dx
dy
,
dx
Solving for
we get
dy
=
dx
1
y
ln(6)6cos(x−y) (sin(x − y)) − x1
2
+ ln(6)6cos(x−y) (sin(x − y)) − ln(5)y
(2) Use logarithmic
to find the derivatives of the following functions:
√ differentiation
x2 + 5(x + 2)7
(a) f (x) =
(cos(x))5/2
√
x2 +5(x+2)7
,
(cos(x))5/2
Setting y =
√
ln(y) = ln(
Applying
d
dx
x2 + 5(x + 2)7
1
5
) = ln(x2 + 5) + 7 ln(x + 2) − ln(cos(x))
5/2
(cos(x))
2
2
to both sides, we get:
dy
dx
y
Solving for
we have:
=
1 1
1
5 1
(2x)
+
7
−
(− sin(x))
2 x2 + 5
x + 2 2 cos(x)
dy
,
dx
we get:
√
dy
x2 + 5(x + 2)7
x
7
5
=
·( 2
+
+ tan(x))
5/2
dx
(cos(x))
x +5 x+2 2
√
(b) g(x) = csc(x)
x3 −7x
√
Setting y = csc(x)
Applying
d
dx
x3 −7x
, we have:
√
ln(y) = x3 − 7x ln(csc(x))
to both sides, we get:
dy
dx
√
1
1
= √
(3x2 − 7) ln(csc(x)) + x3 − 7x
(− csc(x) cot(x))
y
csc(x)
2 x3 − 7x
Solving for
dy
,
dx
we get:
√
dy
(3x2 − 7) ln(csc(x)) √ 3
3
√
− x − 7x cot(x))
= csc(x) x −7x · (
dx
2 x3 − 7x
(c) h(x) = ln(x2 − 7)arctan(x
2 +5x)
Setting y = ln(x2 − 7)arctan(x
2 +5x)
, we have:
ln(y) = arctan(x2 + 5x) ln(ln(x2 − 7x))
Applying
dy
dx
y
=
d
dx
to both sides, we get:
1
1
1
2
2
(2x
+
5)
ln(ln(x
−
7x))
+
arctan(x
+
5x)
·
· (2x − 7)
(x2 + 5x)2 + 1
ln(x2 − 7x) x2 − 7x
Solving for
dy
,
dx
we get:
dy
(2x + 5) ln(ln(x2 − 7x)) (2x − 7) arctan(x2 + 5x)
=
+
dx
(x2 + 5x)2 + 1
ln(x2 − 7x)(x2 − 7x)
(3) We discussed in class how to find the derivatives of arcsin(x) and arctan(x). Use a similar
method to find the derivatives of:
(a) arccos(x)
dy
Set y = arccos(x). Then cos(y) = x. Differentiating both sides, we get − sin(y) dx
=
√
dy
1
1
2
1, so dx = − sin(y) = − sin(arccos(x)) . Using a right triangle with sides x, 1 − x , and
√
hypotenuse 1, we can conclude that sin(arccos(x)) = 1 − x2 . We then have that
d
1
arccos(x) = − √1−x
2.
dx
(b) arccot(x)
dy
Set y = arccot(x) Then cot(y) = x. Differentiating both sides, we get − csc2 (y) dx
=
dy
2
2
1, so dx = − sin (y) = − sin (arccot(x)). Using a right triangle with sides x, 1, and
√
hypotenuse x2 + 1, we can conclude that sin(arccot(x)) = √x12 +1 . We then have
d
that dx
arccot(x) = − x21+1 .
(c) arcsec(x)
dy
Set y = arcsec(x). Then sec(y) = x. Differentiating both sides, we get sec(y) tan(y) dx
=
dy
1, so dx = cos(y) cot(y) = cos(arcsec(x)) cot(arcsec(x)). Using a right triangle with
√
sides 1, x2 − 1, and hypotenuse x, we can conclude that cos(arcsec(x)) = x1 and
cot(arcsec(x)) = √x12 −1 . There is a small wrinkle here: sec(arcsec(x)) tan(arcsec(x))
must always be positive. The range of arcsec(x) is [0, π]. So, in the first quadrant, sec(arcsec(x)) and tan(arcsec(x)) and, in the
√ second quadrant, sec(arcsec(x))
and tan(arcsec(x)) are both negative. So, since x2 − 1 is always non-negative, we
must take |x| instead of x to ensure we have the correct sign. So, we have that
d
arcsec(x) = |x|√1x2 −1 .
dx
(d) arccsc(x)
dy
Set y = arccsc(x). Then csc(y) = x. Differentiating both sides, we get − csc(y) cot(y) dx
=
dy
1, so dx = − sin(y) tan(y) = − sin(arccsc(x)) tan(arccsc(x)). Using a right triangle
√
with sides x2 − 1, 1, and hypotenuse x, we can conclude that sin(arccsc(x)) = x1
and tan(arccsc(x)) = √x12 −1 . For the same reasoning as above, we need to take |x|
d
arccos(x) = − |x|√1x2 −1 .
instead of x in the denominator. We then have that dx
(4) The graph of the of the curve (x2 + y 2 − 2x)2 = 4(x2 + y 2 ), known as a cardioid, is given
below.
dy
(a) Use implicit differentiation to find dx
.
Applying
d
dx
to both sides of (x2 + y 2 − 2x)2 = 4(x2 + y 2 ), we get:
2(x2 + y 2 − 2x)(2x + 2y
Solving for
dy
,
dx
dy
dy
− 2) = 4(2x + 2y )
dx
dx
we get:
dy
x3 + xy 2 − 3x2 − y 2
=
dx
y(2 + 2x − x2 − y 2 )
dy
(b) When does dx
not exist? Use your answer from part a to figure out where
doesn’t make sense (i.e., by dividing by zero).
dy
dx
dy
dx
literally
doesn’t exist when y(2 + 2x − x2 − y 2 ) = 0. So, this happens when y = 0 or
when x2 + y 2 − 2x − 2 = 0. To work with x2 + y 2 − 2x − 2 = 0, we use the fact that
we want pairs (x, y) that satisfy both x2 + y 2 − 2x − 2 = 0 as well as
p our original
2
2
equation. The original equation to tell us that x + y − 2x =p±2 x2 + y 2 . So
x2 + y 2 − 2x − 2 = 0 plus this extra information tells us that ±2 x2 + y 2 = 2. We
can conclude that we need x2 + y 2 = 1.
So now, we need to find the intersection of x2 +y 2 = 1 and (x2 +y 2 −2x)2 = 4(x2 +y 2 ).
Plugging in the first equation to the second, we have (1−2x)2 = 4. Then 1−2x = ±2,
meaning x = − 21 or x = 23 . But x = 32 doesn’t make sense with the equation
2
x2 + y√
= 1, so we must have x = − 12 . Again, the equationx2 + y 2 = 1 tells us
y = ± 23 .
√
dy
In conclusion, dx
is not defined for (x, y) pairs (0, 0), (4, 0), (−1/2, 3/2), and
√
(−1/2, − 3/2).
(c) Plot some of the points you found in part b on the graph below (specifically, when
dy
at these points using the graph?
y = 0). Can you explain the non-existence of dx
When y = 0, the curve has a “corner-like” point (at (0, 0)) and a point where
the graph
tangent line (at (4, 0)). For y 6= 0, the points
√ should have a vertical
√
(−1/2, 3/2) and (−1/2, − 3/2) correspond to the two left-most points of the
graph where the graph should also have vertical tangent lines.