PSL Course Packet
MATH 110
Implicit Differentiation
& Related Rates
Covering Material from Section 3.6
Produced in collaboration with the Penn State
Department of Mathematics
Printed copies of this packet are available for FREE from Penn State Learning
c The Pennsylvania State University 2018. This material is not licensed for resale.
2
Implicit Differentiation
Explicit versus Implicit Functions
Definition 1. Suppose we have an equation relating two variables, x and y. If y is
isolated on one side of the equation, we say that y is explicitly defined as a function
of x, or simply, y is a function of x. If y is not isolated on one side of the equation,
we say that y is defined implicitly in terms of x, or y is an implicit function of x.
Explicit Functions
Implicit Functions
• x2 + xy − y = 1
• y = x2 − 3
√
• y = x+7
• x3 + y 3 + yx − 4 = 0
Computing Derivatives Implicitly
When computing the derivative of a function involving x and y (where we treat y
as a function of x), it may be helpful to replace y with f (x) and recall the general
power rule:
d
[f (x)]n = n[f (x)]n−1 f 0 (x)
dx
Example 1. Compute the derivative of x2 y 3 with respect to x by treating y as a
function of x.
Step 1. Replace y with f (x).
Since we are treating y as a function of x, we can rewrite x2 y 3 as
x2 [f (x)]3 .
Step 2. Differentiate with respect to x.
Since x2 [f (x)]3 is a product of two functions of x, namely x2 and [f (x)]3 , we begin
by using the product rule.
d 2
d
d 2
3
3
2
3
x [f (x)] =
x [f (x)] + x
[f (x)]
Product Rule
dx
dx
dx
= 2x[f (x)]3 + x2 3[f (x)]2 f 0 (x)
Power & General Power rules
Step 3. And finally, replace f (x) with y and f 0 (x) with y 0 .
d 2 3
x y = 2xy 3 + 3x2 y 2 y 0
dx
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3
The Method of Implicit Differentiation
Suppose we are given an equation relating the variables x and y. We can compute
the derivative of y with respect to x using the following technique.
1. Differentiate both sides of the equation with respect to x.
2. If given, plug in specific values of x and y into resulting equation.
3. Solve for y 0 in terms of x and y.
(a) Put all terms with a factor of y 0 on the left side of the equation and
everything else on the right side of the equation.
(b) Factor out y 0 from each term on the left side of the equation.
(c) Divide both sides by the appropriate factor to solve for y 0 .
Example 2. Find
dy
where y is defined implicitly by x2 y − x4 + y 3 = 1.
dx
Step 1. Replace y with f (x).
x2 f (x) − x4 + [f (x)]3 = 1
Step 2. Differentiate both sides with respect to x.
[2xf (x) + x2 f 0 (x)] − 4x3 + 3[f (x)]2 f 0 (x) = 0
Notice the use of the product rule to compute the derivative of x2 f (x) and the
general power rule to compute the derivative of [f (x)]3 .
Step 3. Rearrange terms so that only terms involving f 0 (x) are on the left side.
x2 f 0 (x) + 3[f (x)]2 f 0 (x) = 4x3 − 2xf (x)
Step 4. Factor out f 0 (x) on the left side of the equation.
(x2 + 3[f (x)]2 )f 0 (x) = 4x3 − 2xf (x)
Step 5. Solve for f 0 (x).
f 0 (x) =
4x3 − 2xf (x)
x2 + 3[f (x)]2
Step 6. And finally, replace f (x) with y and f 0 (x) with y 0 (or
dy
).
dx
dy
4x3 − 2xy
= 2
dx
x + 3y 2
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4
Observation
Once we get used to the idea of thinking of y as a function of x, it’s not necessary
to replace y with f (x). Instead, we can use the general power rule written in the
following form:
d n
y = ny n−1 y 0
dx
dy
= ny n−1
dx
Example 3. Find the equation of the line tangent to the curve defined implicitly
by
x2 y 3 − y 2 + xy = 5
at the point (2, 1).
Step 1. Differentiate both sides with respect to x.
dy
dy
3
2
2 dy
+ y+x
− 2y
=0
2xy + x 3y
dx
dx
dx
See Example 1 for how to differentiate x2 y 3 implicitly.
Step 2. Plug in specific values of x and y.
Since we want to find the slope of the line tangent to the curve at the point (2, 1),
plug in x = 2 and y = 1.
4 + 12
dy
dy
dy
−2 +1+2
=0
dx
dx
dx
Step 3. Simplify and rearrange terms.
12
Step 4. Solve for
dy
.
dx
dy
= −5
dx
dy
5
=−
dx
12
Step 5. Use the point-slope equation of a line: y − b = m(x − a).
The equation of the line tangent of the curve at the point (2, 1) is
y−1=−
5
(x − 2).
12
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5
Example 4. Find
dy
at (2, 0) where y is defined implicitly by
dx
p
xy + x = 2x + 7y.
Step 1. Differentiate both sides with respect to x.
d
dy
(xy + x) = y + x + 1
dx
dx
dp
2x + 7y
dx
=
=
=
d
(2x + 7y)1/2
dx
1
d
(2x + 7y)−1/2 (2x + 7y)
2
dx
1
dy
√
2+7
dx
2 2x + 7y
Therefore,
dy
1
y+x +1= √
dx
2 2x + 7y
dy
2+7
dx
Step 2. Plug in the values x = 2 and y = 0.
dy
1
dy
2 +1=
2+7
dx
4
dx
Step 3. Expand both sides of the equation so that each side is written as a sum.
2
dy
1 7 dy
+1= +
dx
2 4 dx
Step 4. Rearrange terms (i.e., subtract 1 from both sides and subtract
both sides).
7 dy
from
4 dx
1 dy
1
=−
4 dx
2
Step 5. Solve for
dy
(i.e., multiply both sides by 4).
dx
dy
= −2
dx
Alternatively, we could multiply both sides of the equation in Step 2 by 4 to get
8
dy
dy
+4=2+7 .
dx
dx
Now rearranging terms is a little easier since we don’t have to deal with any fractions.
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6
Example 5. The demand equation for Dr. Hager’s Make Math Great Again video
series is given by
p = −4x2 − 2x + 30
where p is the wholesale unit price in dollars and x is the quantity demanded in
units of a thousand. Compute the elasticity of demand when x = 2.
Observation: Ordinarily when we compute elasticity of demand, the first thing we
do is find x = f (p) by solving for x in terms of p in the given demand equation.
However in this case, solving for x is easier said than done. So instead, we will use
implicit differentiation to compute f 0 (p) when the time comes.
Step 1. Compute the unit price p when x = 2.
p = −4(22 ) − 2(2) + 30
= −16 − 4 + 30
= 10
since p = −4x2 − 2x + 30
Step 2. Compute f (10).
Recall that x = f (p) is the quantity demanded (in thousands) when the unit price
is p dollars. We were given x = 2 (and used that to determine p = 10 in Step 1),
which is equivalent to saying f (10) = 2.
Step 3. Compute f 0 (10).
Compute the derivative of both sides of the demand equation (p = −4x2 − 2x + 30)
with respect to p. Here is where we use implicit differentiation by thinking of x as
a function of p (i.e., x = f (p)).
d
(−4x2 − 2x + 30)
dp
= −8xx0 − 2x0
1=
since
d
p=1
dp
Now plug in x = 2 to get
1 = −16x0 − 2x0 = −18x0
and therefore x0 = f 0 (10) = −1/18.
Step 4. Compute elasticity of demand using the values p = 10, f (10) = 2, and
f 0 (10) = −1/18.
10 · (−1/18)
E(10) = −
2
10 1
=
·
2 18
5
=
18
p · f 0 (p)
since E(p) = −
f (p)
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7
Related Rates
How to Solve a Related Rates Problem
Here are a few guidelines to follow when solving a related rates problem.
1. Assign a variable to each quantity that changes and state the known and
unknown rates of change with respect to time t.
2. Find an equation relating the variables. Don’t forget about business models.
3. Differentiate both sides of the equation implicitly with respect to t. Treat all
variables as functions of t.
4. Replace the variables and their derivatives by the numerical data given. Watch
out for words like “increasing” or “decreasing”, “rising” or “falling”, “away” or
“towards”, etc. to determine if a derivative is positive or negative.
5. Solve the equation for the unknown rate of change.
Example 6. Given p2 x + 6x = 5 and
dp
dx
= 3. Compute
when p = 2.
dt
dt
Step 1. Assign variables and state the known and unknown rates of change.
Variables (as functions of t): x(t) and p(t)
dp
Known rate of change:
=3
dt
dx
Unknown rate of change:
when p = 2
dt
Step 2. Find an equation which relates the variables.
p2 x + 6x = 5
Step 3. Differentiate both sides of the equation implicitly with respect to t.
d 2
(p x + 6x) = 2pp0 x + p2 x0 + 6x0
dt
Implicit differentiation & product rule
d
5=0
dt
Step 4. Since x appears in the equation after differentiation, we need to find its
value by plugging in p = 2 into the given equation p2 x + 6x = 5.
=0
(2)2 x + 6x = 5
since
⇒
10x = 5
⇒
x = 1/2
Step 5. Plug in the known values of the variables and their derivatives into the
equation found in Step 3. (p = 2, x = 1/2, dp
= 3)
dt
1
2(2)(3) + (2)2 x0 + 6x0 = 0
⇒
6 + 10x0 = 0
2
dx
dp
Therefore
= −3/5 when p = 2 and
= 3.
dt
dt
⇒
x0 = −3/5
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8
Example 7. A company is decreasing the production of energy drinks at a rate of
100 cases per day. All cases produced can be sold. The daily demand function is
given by
x
p(x) = 50 −
200
where x is the number of cases produced and sold, and p(x) is the unit price in
dollars. Find the rate of change of the revenue with respect to the time in days
when daily production is 300 cases.
Step 1. Assign variables and state the known and unknown rates of change.
Variables (as functions of t): x(t) = the quantity produced and sold per day
R(t) = the daily revenue
Known rate of change:
Unknown rate of change:
dx
= −100
dt
dR
dt
since production is decreasing
when x = 300
Step 2. Find an equation which relates the variables.
R = x · p(x)
x = x 50 −
200
1 2
= 50x −
x
200
Revenue Function
since p(x) = 50 −
x
200
Simplify
Step 3. Differentiate both sides of the equation implicitly with respect to t.
1 2
50x −
x
200
dx
1
dx
−
2x
= 50
dt
200
dt
dx
x dx
= 50
−
dt
100 dt
dR
d
=
dt
dt
Implicit differentiation & power rule
Simplify
Step 4. Plug in the known values of the variables and their derivatives into the
equation found in Step 3. (x = 300, dx
= −100)
dt
dR
300
= 50(−100) −
(−100)
dt
100
= −5000 + 300
= −4700
Therefore revenue is decreasing at $4700 per day when the level of production is 300
cases per day and decreasing at 100 cases per day.
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9
Example 8. The weekly demand function is given by
2p + x + 4xp = 22
where x is the number of thousands of units demanded weekly and p is in dollars.
If the price is increasing at a rate of 25 cents per week when the weekly demand is
4000 units, at what rate is the demand decreasing?
Step 1. Assign variables and state the known and unknown rates of change.
Variables (as functions of t): x(t) = demand in units of a thousand
p(t) = unit price in dollars
Known rate of change:
25
dp
=
dt
100
since price (in dollars) is increasing
dx
when x = 4
dt
Step 2. Find an equation which relates the variables.
Unknown rate of change:
2p + x + 4xp = 22
Step 3. Differentiate both sides of the equation implicitly with respect to t.
dp dx
dx
dp
d
(2p + x + 4xp) = 2 +
+ 4 p + 4x
Implicit differentiation & product rule
dt
dt
dt
dt
dt
d
=0
since 22 = 0
dt
Step 4. Since p appears in the equation after differentiation, we need to find its
value by plugging in x = 4 into the given demand equation 2p + x + 4xp = 22.
2p + 4 + 4(4)p = 22
⇒
18p + 4 = 22
⇒
p=1
Step 5. Plug in the known values of the variables and their derivatives into the
equation found in Step 3. (x = 4, p = 1, dp
= 25/100)
dt
0=2
1
2
9
=
2
=
25
dx
dx
25
+
+ 4 (1) + 4(4)
100
dt
dt
100
dx
+5 +4
dt
dx
+5
dt
Simplify
Therefore
9
dx
dx
9
dx
9
+5
=0
⇒
5
=−
⇒
=−
2
dt
dt
2
dt
10
and since x is measured in thousands of units, this means that demand is decreasing
by 900 units per week (i.e., 9/10 of one thousand) when the weekly demand is 4000
units and the price is increasing at 25 cents per week.
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10
Example 9. The wholesale price p of an e-tablet writing stylus in dollars is related
to the supply x in thousands by:
125p2 − x2 = 100
If 20,000 styluses are available at the beginning of a week, and the price is falling at
4 cents per week, at what rate is the supply falling?
Step 1. Assign variables and state the known and unknown rates of change.
Variables (as functions of t):
Known rate of change:
Unknown rate of change:
x(t) = supply in units of a thousand
p(t) = unit price in dollars
4
dp
=−
dt
100
dx
dt
since price (in dollars) is falling
when x = 20
Step 2. Find an equation which relates the variables.
125p2 − x2 = 100
Step 3. Differentiate both sides of the equation implicitly with respect to t.
250p
dp
dx
− 2x
=0
dt
dt
Power rule
Step 4. Since p appears in the equation after differentiation, we need to find its
value by plugging in x = 20 into the given supply equation 125p2 − x2 = 100.
125p2 − 202 = 100
⇒
125p2 = 500
⇒
p2 = 4
⇒
p=2
Step 5. Plug in the known values of the variables and their derivatives into the
equation found in Step 3. (x = 20, p = 2, dp
= −4/100)
dt
4
dx
0 = 250(2) −
− 2(20)
100
dt
4
dx
= −500
− 40
100
dt
dx
= −20 − 40
dt
Simplify
Therefore
−20 − 40
dx
=0
dt
⇒
−40
dx
= 20
dt
⇒
dx
1
=−
dt
2
and since x is measured in thousands of units, this means that supply is falling by
500 units per week (i.e., 1/2 of one thousand) when the weekly demand is 20,000
styluses and the price is falling at 4 cents per week.
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11
Try It Yourself
Exercise 1. Find
dy
where y is defined implicitly by y 4 + x3 y 2 − x = 5.
dx
Exercise 2. Find the equation of the line tangent to the curve defined implicitly
by
x4 y − x3 + 14y = 29
at the point (1, 2).
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12
Exercise 3. The demand equation for Hager & Little’s Better Business Calculus
book series is given by
p = −5x2 − 2x + 100
where p is the wholesale unit price in dollars and x is the quantity demanded in
units of a thousand. Compute the elasticity of demand when x = 4.
Exercise 4. A company is increasing production of math-brain protein bars at a
rate of 50 cases per day. All cases produced can be sold. The daily demand function
is given by
x
p(x) = 80 −
100
Where x is the number of cases produced and sold, and p is in dollars. Find the rate
of change of the revenue with respect to the time in days when daily production is
500 cases.
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13
Exercise 5. If the quantity demanded daily of a product is related to its unit price
in dollars by
p2 = 106 − x2
How fast is the quantity demanded changing when x = 5 and the unit price is
decreasing at a rate of $3 per day.
Exercise 6. The weekly demand function is given by
p + x + 3xp = 43
where x is the number of thousands of units demanded weekly and p is in dollars.
If the price is increasing at a rate of 20 cents per week when the level of demand is
4000 units, at what rate is the demand decreasing?
Answers: 1) (1 − 3x2 y 2 )/(4y 3 + 2x3 y) 2) y = −x/3 + 7/3 3) 1/14
4) revenue increases by $3,500 per day 5) demand increases by 27/5 units per day
6) demand decreases by 260 units per week
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