AB
College Board AP® Calculus
Your notes
Behaviors of Implicit Relations
Contents
Second Derivatives of Implicit Functions
Critical Points of Implicit Relations
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1
Second Derivatives of Implicit Functions
Second derivatives of implicit functions
Your notes
How do I find the second derivative of an implicit
function?
First make sure you are comfortable with the content covered in the 'Implicit
Differentiation' study guide
This involves being able to find the derivative of a function defined implicitly, using
the chain rule
We are able to find the second derivative,
the first derivative,
d 2y
, by differentiating the expression for
dx 2
dy
dx
Consider finding the second derivative of x 2
+ y 2 = 4x
Find the first derivative, this is shown in the Implicit Differentiation study guide to be
dy 2 − x
= y
dx
To find the second derivative, differentiate both sides with respect to x
d ⎛⎜ dy ⎞⎟ d ⎛⎜ 2 − x ⎞⎟
⎜
⎟=
⎜
⎟
dx ⎝ dx ⎠ dx ⎝ y ⎠
The derivative of
dy
d 2y
with respect to x is
dx
dx 2
d 2y
d ⎛⎜ 2 − x ⎞⎟
⎜
⎟
=
dx 2 dx ⎝ y ⎠
The right hand side can then be differentiated with respect to x , in this case using the
u ⎞⎟ ' u 'v − uv '
⎟ =
v⎠
v2
⎝
⎛
quotient rule, ⎜⎜
u = 2 − x and v = y
Differentiate u and v with respect to x , remember to apply the chain rule when
differentiating y
u ' = − 1 and v ' =
dy
dx
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2
Applying the quotient rule to the right hand side
dy
d 2y
=
dx 2
Your notes
−1 · y − 2 − x · dx
(
)
y2
If your answer references the first derivative, substitute it in
dy 2 − x
= y
dx
We know from before that
d 2y
dx 2
=
−1 · y − 2 − x ·
(
)
2 − x ⎞⎟
⎟
y ⎠
⎝
⎛
⎜
⎜
y2
This is the correct second derivative, but it can be simplified
d 2y
=
dx 2
−y −
(
2−x 2
y
)
y2
Writing with a common denominator can help simplify
d 2y −y
2 − x 2 −y 2
2−x 2
= y2 − y3 = y3 − y3
dx 2
(
)
(
)
d 2y
y2 + 2 − x 2
=
−
dx 2
y3
(
)
This is the final answer, but expressions like this can be written in several different ways,
E.g. by factoring out negative signs in a different place
Examiner Tips and Tricks
Remember that:
dy
dx
dy
d 2y
The derivative of
with respect to x is
dx
dx 2
The derivative of y with respect to x is
Worked Example
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3
+ cos y = 0 . 8 can be written as:
sin x sin2y + cos2x cos y
Show that the second derivative of sin x
d 2y
=−
dx 2
Your notes
sin3y
Answer:
Find the first derivative by differentiating both sides with respect to x
d
d
sin x + cos y =
0.8
dx
dx
d
d
d
sin x +
cos y =
0.8
dx
dx
dx
(
)
(
)
(
)
Remember to use the chain rule when differentiating cos y with respect to x
cos x + − sin y ·
dy
=0
dx
dy cos x
= sin y
dx
This could instead be simplified to cos x csc y , depending on whether you would
prefer to apply the quotient rule to
cos x
or the product rule to cos x csc y for
sin y
the next part
To find the second derivative, differentiate both sides with respect to x again
d ⎛⎜ dy ⎞⎟ d ⎛⎜ cos x ⎞⎟
⎜
⎟=
⎜
⎟
dx ⎝ dx ⎠ dx ⎝ sin y ⎠
d 2y
d ⎛⎜ cos x ⎞⎟
⎜
⎟
=
dx 2 dx ⎝ sin y ⎠
Apply the quotient rule to the right hand side, remember to apply the chain rule when
differentiating sin y with respect to x
u ⎞⎟ ' u 'v − uv '
⎟ =
v⎠
v2
⎝
⎛
⎜
⎜
u = cos x v = sin y
u ' = − sin x v ' = cos y ·
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dy
dx
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4
dy
d 2y
=
dx 2
−sin x · sin y − cos x · cos y · dx
Substitute in the expression for
d 2y
=
dx 2
Your notes
sin2y
dy
dy cos x
= sin y
found previously,
dx
dx
cos x ⎞⎟
−sin x · sin y − cos x · cos y · sin y
⎛
⎜
⎜
⎝
⎟
⎠
sin2y
Now we need to rearrange into the form given in the question
Write as two fractions to see if they can be simplified
d 2y −sin x sin y cos2x cos y
= sin2y − sin3y
dx 2
They can be written with a common denominator
d 2y −sin x sin2y cos2x cos y
= sin3y − sin3y
dx 2
d 2y −sin x sin2y − cos2x cos y
=
dx 2
sin3y
Put the negative at the front of the entire fraction, to match the form given in the
question
d 2y
sin x sin2y + cos2x cos y
=−
dx 2
sin3y
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5
Critical Points of Implicit Relations
Your notes
Critical points of implicit relations
Do implicit equations have critical points?
Equations defined implicitly can have critical points
Critical points are defined in the same way as they are for any other function
A critical point occurs where the derivative is equal to zero
or where the derivative does not exist
The applications of first and second derivatives to classify the nature of points on a
graph can be extended to implicit functions
These properties are summarized in the table below
Type of point
First derivative
Second derivative
Local minimum
Zero
Positive or zero
Local maximum
Zero
Negative or zero
Point of inflection (critical)
Zero
Zero
Point of inflection (non-critical)
Non-zero
Zero
Remember that second derivative equal to zero is not enough for a point to be a point of
inflection
The second derivative must change sign at the point as well
How can I find points on an implicitly-defined curve
where the tangent is horizontal or vertical?
The tangent line to an implicitly-defined curve will be horizontal at a point on the curve
where
dy
=0
dx
The tangent line to an implicitly-defined curve will be vertical at a point on the curve
where
dx
=0
dy
This is the same as a point at which
dy
has a denominator equal to zero
dx
and a numerator not equal to zero
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6
Recall
dx
1
=
dy ⎛⎜ dy ⎞⎟
⎜
⎟
dx ⎠
⎝
Worked Example
=
Your notes
Consider the function y f x whose curve is given by the equation
2y 2 6 y sin 2x for y 0 .
− =
(a) For 0
(
>
)
≤ x ≤ π2 and y > 0, find the coordinates of the point where the tangent to
the curve is horizontal.
Answer:
We need to find where the derivative is equal to zero, as the slope of the tangent is
zero at that point
Differentiate both sides of the equation with respect to x
Use the product rule for the right hand side, and don't forget to use the chain rule
when differentiating y with respect to x
d
d
2y 2 − 6 =
y sin 2x
dx
dx
dy dy
4y ·
= · sin 2x + y · 2cos 2x
dx dx
(
Rearrange for
)
(
)
dy
dx
4y ·
dy dy
− · sin 2x = 2y cos 2x
dx dx
dy
4y − sin 2x = 2y cos 2x
dx
dy
2y cos 2x
=
dx 4y − sin 2x
(
)
Set this equal to zero to find the critical point
2y cos 2x
=0
4y − sin 2x
2y cos 2x = 0
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We also need to be careful with the denominator, so that the derivative is not
undefined (from dividing by zero)
− sin 2x ≠ 0
Solve to find x , and use the fact that we know y > 0 , so 2y cos 2x = 0 will have the
same solutions as cos 2x = 0
2y cos 2x = 0
cos 2x = 0
π 3π , ...
2x = ,
and 4y
Your notes
2
2
π 3π , ...
x= ,
4 4
We were told in the question that 0
≤ x ≤ π2
x=
π
4
− 6 = y sin 2x
π
2y 2 − 6 = y sin 2 ·
Find the y value by substituting into 2y 2
⎛
⎜
⎝
2y 2 − 6 = y sin ⎛⎜
2y 2 − 6 = y
(
π
⎞
⎟
4 ⎠
⎞
⎟
⎝ 2 ⎠
2y 2 − y − 6 = 0
2y + 3 y − 2 = 0
) (
)
y > 0 , so y = 2
Check this satisfies 4y
− sin 2x ≠ 0
4 2 − sin ⎛⎜2 ·
(
)
⎝
π =8−1=7≠0
⎞
⎟
4 ⎠
⎛
The point where the tangent to the curve is horizontal is ⎜
π,2
⎝ 4
⎞
⎟
⎠
(b) Determine whether the curve has a relative minimum, a relative maximum, or
neither at the point found in part (a). Justify your answer.
Answer:
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⎛
We need to check if the second derivative is positive or negative at ⎜
⎝
π,2
4
⎞
⎟
⎠
Your notes
Write down the first derivative, and then differentiate both sides with respect to x
dy
2y cos 2x
=
dx 4y − sin 2x
d 2y
d ⎛⎜ 2y cos 2x ⎞⎟
⎜
⎟
=
dx 2 dx ⎝ 4y − sin 2x ⎠
u ⎞⎟ ' u 'v − uv '
⎟ =
with u = 2y cos 2x and
v⎠
v2
⎝
⎛
Use the quotient rule, ⎜⎜
v = 4y − sin 2x
Differentiating v is relatively straightforward using the chain rule
v = 4y − sin 2x
dy
v'=4
− 2cos 2x
dx
Differentiating u will require the product rule
Use different variables than u and v so you don't get confused, e.g. p and q
u = 2y cos 2x
p = 2y q = cos 2x
p'=2
dy
q ' = − 2sin 2x
dx
u'=p' q + p q'
dy
· cos 2x + 2y · − 2sin 2x
dx
dy
u ' = 2cos 2x ·
− 4y sin 2x
dx
u'=2
'
u ⎞⎟ ' u v − uv
⎟ =
v⎠
v2
⎝
⎛
Apply the quotient rule, ⎜⎜
d 2y
dx 2
=
⎛
⎜
⎜
⎝
2cos2x ·
'
dy
dy
⎞
⎛
⎞
−
4y sin2x ⎟⎟ · 4y − sin2x − 2y cos2x · ⎜⎜4
−
2cos2x ⎟⎟
dx
dx
⎠
⎝
⎠
2
4y − sin2x
(
)
(
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This is quite a tricky expression, but thankfully we do not need to rearrange it, or even
substitute in the expression
⎛
We know that the point is ⎜
⎝
dy
dx
Your notes
π , 2 and we know that at this point, dy = 0
⎞
⎟
4
dx
⎠
dy
d 2y
So we can substitute these values of x , y , and
into the expression for
dx
dx 2
d 2y
dx 2
=
⎛
⎜
⎝
0 − 4 2 sin ⎛⎜2 ·
(
)
⎝
π · 4 2 − sin 2 · π − 2 2 cos 2 · π · 0 − 2cos 2 · π
4
4
4
4
⎞⎞
⎟⎟
(
(
)
(
))
⎠⎠
(
(
π
4 2 − sin 2 · 4
(
(
)
)
(
(
))
(
(
))
2
))
d 2y
−8 · 8 − 1 − 0 · 0
56
=
=
−
49
dx 2
8−1 2
(
)
(
)
(
(
)
(
)
)
Use the sign of the second derivative to classify the nature of the critical point
⎛
⎜
⎝
π,2
4
⎛
At ⎜
⎞
⎟
⎠
π , 2 , the first derivative is zero, and the second derivative is negative.
⎝ 4
⎞
⎟
⎠
Therefore it is a relative maximum.
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