9/19/12
Worksheet: Implicit Diff., Higher Derivatives
Exercise 1. Use implicit differentiation to find y 0 =
dy
dx
GSI: Ralph Morrison
for the following equations.
• x3 + 2y 2 = 4xy 3 .
Solution. Differentiating both sides gives
3x2 + 4yy 0 = 4y 3 + 12xy 2 y 0 .
Separating out the y 0 terms gives
4yy 0 − 12xy 2 y 0 = 4y 3 − 3x2 ,
or
(4y − 12xy 2 )y 0 = 4y 3 − 3x2 .
Dividing gives us our solution:
y0 =
4y 3 − 3x2
.
4y − 12xy 2
• sin(x/y) = 3x − 2 arctan(y 2 ).
Solution. Differentiating both sides gives
cos(x/y) ·
1
y − xy 0
=3−2
· 2yy 0
2
y
1 + (y 2 )2
Separating out the y 0 terms gives
−
or
x
4y
1
cos(x/y)y 0 +
· y 0 = 3 − cos(x/y),
2
4
y
1+y
y
x
4y
− 2 cos(x/y) +
y
1 + y4
y0 = 3 −
1
cos(x/y).
y
Dividing gives us our solution:
0
y =
3 − y1 cos(x/y)
− yx2 cos(x/y) +
4y
1+y 4
.
• ex+y − ln(x) = y 2 .
Math 10A: Methods of Mathematics
1
Fall 2012
9/19/12
Worksheet: Implicit Diff., Higher Derivatives
GSI: Ralph Morrison
Solution. Differentiating both sides gives
ex+y (1 + y 0 ) −
1
= 2yy 0
x
Separating out the y 0 terms gives
1
ex+y y 0 − 2yy 0 = −ex+y + ,
x
or
1
(ex+y − 2y)y 0 = −ex+y + .
x
Dividing gives us our solution:
−ex+y + x1
y = x+y
.
e
− 2y
0
Exercise 2. As we saw in lecture, the first, second, and third derivatives of position with respect
to time are called velocity, acceleration, and jerk. The fourth, fifth and sixth derivatives of position with respect to time are called snap, crackle, and pop. (Like, for real. Wikipedia the word
“jounce.”)
• What are the snap, crackle, and pop if position is given by f (x) = sin(x)? What about
f (x) = sin(2x)?
Solution. The function f (x) = sin(x) follows a very nice derivative pattern, repeating every
four derivatives:
f 0 (x) = cos(x)
f 00 (x) = − sin(x)
f 000 (x) = − cos(x)
f (4) (x) = sin(x)
f (5) (x) = cos(x)
f (6) (x) = − sin(x).
So the snap, crackle, and pop are f (4) (x) = sin(x), f (5) (x) = cos(x), and f (6) (x) =
− sin(x). If we’d considered f (x) = sin(2x) instead, we’d have the same derivatives except
each time the chain rule would make us multiply by 2; for instance, f 0 (x) = 2 cos(2x). Thus
we’d have f (4) (x) = 24 sin(2x), f (5) (x) = 25 cos(2x), and f (6) (x) = −26 sin(2x).
• Suppose position is given by g(x) = xn for some positive n. If n ≥ 6, what are the snap,
crackle, and pop? What about if n < 4? (What about for n = 4 and n = 5?)
Math 10A: Methods of Mathematics
2
Fall 2012
9/19/12
Worksheet: Implicit Diff., Higher Derivatives
GSI: Ralph Morrison
Solution. When we take derivatives of g, we find things like
g 0 (x) = nxn−1
g 00 (x) = n(n − 1)xn−2
and so on. Thus we’ll have
g (4) (x) = n(n − 1)(n − 2)(n − 3)xn−4 =
n!
xn−4
(n − 4)!
g (5) (x) = n(n − 1)(n − 2)(n − 3)(n − 4)xn−5 =
n!
xn−5
(n − 5)!
g (6) (x) = n(n − 1)(n − 2)(n − 3)(n − 4)(n − 5)xn−6 =
n!
xn−6 ,
(n − 6)!
at least if n ≥ 6. If n < 4, then all those derivatives will be 0. If n = 4, then g (5) (x) = 0 and
g (6) (x) = 0. If n = 5, then g (6) (x) = 0.
2
2
dy
Exercise 3. The equation for a particular ellipse is given by x9 + y4 = 1. Find dx
, and evaluate
√
√
it at the points (2, −2 5/3), (2, 2 5/3), and (3, 0). What’s the geometric interpretation of what
you’ve calculated?
Solution. Differentiating both sides of
x2
9
+
y2
4
= 1 gives us
2x 2y 0
+ y = 0,
9
4
which gives us
y0 = −
2x/9
4x
=− .
y/2
9y
Plugging in our points gives
√
y 0 (2, −2 5/3) = −
√
y 0 (2, 2 5/3) = −
Math 10A: Methods of Mathematics
4·2
4
√
= √ ,
9 · −2 5/3
3 5
4·2
4
√
=− √ ,
9 · 2 5/3
3 5
3
Fall 2012
9/19/12
Worksheet: Implicit Diff., Higher Derivatives
and
y 0 (3, 0) =
GSI: Ralph Morrison
−4 · 3
,
9·0
the last of which is undefined.
√
0
The first
lines tangent to the ellipse at the points (2, −2 5/3)
√ two values of y are the slopes of the
and (2, 2 5/3), respectively. The fact that y 0 (3, 0) does not exist means that for some reason the
derivative does not make sense at (3, 0). Just from this, we don’t know if this is because there is a
corner there or some other issue. However, considering the graph, we see that the tangent line at
this point is vertical, which of course makes the slope undefined.
Math 10A: Methods of Mathematics
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Fall 2012