Sec 2.6 – Marginals and Differentials 2012 Pearson Education, Inc. All rights reserved Let C(x), R(x), and P(x) represent, respectively, the total cost,

Sec 2.6 – Marginals and Differentials

2012 Pearson Education, Inc. All rights reserved

Let C(x), R(x), and P(x) represent, respectively, the total cost, revenue, and profit from the production and sale of x items.

The marginal cost at x, given by C (x), is the approximate cost of the (x + 1)th item:

C (x) ≈ C(x + 1) – C(x), or C(x + 1) ≈ C(x) + C (x).

The marginal revenue at x, given by R (x), is the approximate revenue from the (x + 1)th item:

R (x) ≈ R(x + 1) – R(x), or R(x + 1) ≈ R(x) + R (x).

The marginal profit at x, given by P (x), is the approximate profit from the (x + 1)th item:

P (x) ≈ P(x + 1) – P(x), or P(x + 1) ≈ P(x) + P (x).


Example 1:

Given the following cost and revenue functions,

C(x) 62x2 27,500 and

R(x) x3 12x2 40x 10

find each of the following:a) Total profit, P(x).b) Total cost, revenue, and profit from the production and sale of 50 units

of the product.c) The marginal cost, revenue, and profit when 50 units are produced and sold.

( )P x ( ) ( )R x C x

( )P x 3 2 212 40 10 (62 27,500)x x x x

( )P x

a)

𝑥3−74 𝑥2+40 𝑥−27,490


b) (50)C 262(50) 27,500 $182,500

(50)R

(50)P

$97,010

$85,490

𝑥3−12 𝑥2+40 𝑥+10

𝑥3−74 𝑥2+40 𝑥−27,490

c) ( )C x 124x(50)C 124(50) $6200

( )R x 23 24 40x x (50)R 23(50) 24(50) 40 $6340

( )P x 23 148 40x x

(50)P 23(50) 148(50) 40 $140

The approximate cost of the 51st unit will be $6200.

The approximate revenue from the sale of the 51st unit will be $6340.

The approximate profit from the 51st unit will be $140.


𝑥+∆ 𝑥𝑥

𝑑𝑦

∆ 𝑥 𝑜𝑟 𝑑 𝑥

∆ 𝑦

𝑓 (𝑥)𝐿(𝑥)

𝑓 (𝑥)

𝑓 (𝑥+∆ 𝑥)

𝑓 (𝑥 )+𝑑𝑦

𝑥+∆ 𝑥−𝑥

∆ 𝑦= 𝑓 (𝑥+∆ 𝑥 )− 𝑓 (𝑥 )

For

∆ 𝑦= 𝑓 (𝑥+∆ 𝑥 )− 𝑓 (𝑥 )∆ 𝑦= 𝑓 ( 4+0.1 )− 𝑓 (4 )∆ 𝑦= (4.1 )2 − ( 4 )2

∆ 𝑦=0.81


𝑥+∆ 𝑥𝑥

𝑑𝑦


∆ 𝑦


𝑓 (𝑥)

𝑓 (𝑥+∆ 𝑥)



𝑓 (𝑥 )=√𝑥=𝑥12

Approximate

𝑥=25 , ∆ 𝑥=2

𝑓 ′ (𝑥 )=12𝑥

− 12

∆ 𝑦 ≈ 𝑓 ′ (𝑥 ) ∙ ∆ 𝑥

∆ 𝑦 ≈ 0.2For f a continuous, differentiable function, and small ∆x.

f (x) y

x and y f (x)x.

∆ 𝑦 ≈12

(25 )− 1

2 ∙ 2

√27 ≈ 𝑓 (25 )+∆ 𝑦

√27 ≈ √25+0.2√27 ≈ 5.2

√27=5.1961524


DEFINITION:For y = f (x), we define

dx, called the differential of x, by dx = ∆xand

dy, called the differential of y, by dy = f (x)dx.

𝑥+∆ 𝑥𝑥

𝑑𝑦


∆ 𝑦


𝑓 (𝑥)

𝑓 (𝑥+∆ 𝑥)



Sec 2.6 – Marginals and DifferentialsExamples

𝑦=4 𝑥2− 9𝑥7

𝑑𝑦=8 𝑥𝑑𝑥−63 𝑥6𝑑𝑥

𝑑𝑦= (8 𝑥− 63𝑥6 )𝑑𝑥

𝑦=( 4 𝑥+𝑥4 ) (3 𝑥+2𝑥2 )

𝑑𝑦= ( 4 𝑥+𝑥4 )(3𝑑𝑥+4 𝑥 𝑑𝑥 )+¿(3 𝑥+2 𝑥2 )( 4𝑑𝑥+4 𝑥3𝑑𝑥 )

𝑑𝑦= [ ( 4 𝑥+𝑥4 ) (3+4 𝑥 )+ (3 𝑥+2 𝑥2 ) (4 +4 𝑥3 ) ]𝑑𝑥

𝑦=𝑥

3 𝑥−1

𝑑𝑦=¿(3𝑥−1 ) (1𝑑𝑥 )−𝑥 (3𝑑𝑥 )

(3 𝑥− 1 )2

𝑑𝑦=(3 𝑥− 1 ) (1 ) −𝑥 (3 )

(3 𝑥−1 )2𝑑𝑥

𝑑𝑦= 𝑓 ′ (𝑥 ) ∙𝑑𝑥

𝑑𝑦= ( 4 𝑥+𝑥4 ) +¿(3 𝑥+2 𝑥2 )( 4+4 𝑥3 )𝑑𝑥

𝑑𝑦=−1

(3 𝑥− 1 )2𝑑𝑥

Sec 2.6 – Marginals and DifferentialsExample: For

a) Find dy.b) Find dy when x = 5 and dx = 0.2.

a) 𝑑𝑦=𝑥 ∙3 (𝑥− 4 )2𝑑𝑥+(𝑥− 4 )3 ∙ 1𝑑𝑥

𝑑𝑦= (3 𝑥 (𝑥− 4 )2+(𝑥− 4 )3 )𝑑𝑥

𝑑𝑦= (3 ∙ 5 (5 −4 )2+(5 − 4 )3 ) ∙ 0.2

𝑑𝑦=3.2b)

An Implicit function is one where the variable “y” can not be easily solved for in terms of only “x”.

Examples:

4 𝑥2=2 𝑦3+4 𝑦

𝑥=√ x+𝑦

3 𝑥2𝑦 2=4 𝑥2− 4 𝑥𝑦

𝑥2√𝑦+𝑦2=1

Sec 2.7 – Implicit Differentiation and Related Rates


To differentiate implicitly :

e) Divide both sides of the equation to isolate dy/dx.

a) Differentiate both sides of the equation with respect to x (or whatever variable you are differentiating with respect to).

b) Apply the rules of differentiation as necessary. Any time an expression involving y is differentiated, dy/dx will be a factor in the result.

c) Isolate all terms with dy/dx as a factor on one side of the equation.

d) Factor out dy/dx.

Differentiating an implicit function will create the derivative () needed to calculate the slope of any tangent on the curve or the rate of change at any value of the independent variable.

Find

6 𝑥𝑑𝑥𝑑𝑥

+¿

3 𝑥2+𝑦2=14

2 𝑦𝑑𝑦𝑑𝑥

=¿0

6 𝑥+2 𝑦𝑑𝑦𝑑𝑥

=0

2 𝑦𝑑𝑦𝑑𝑥

=−6 x

𝑑𝑦𝑑𝑥

=−6 𝑥2 𝑦

𝑑𝑦𝑑𝑥

=−3 𝑥𝑦


Find the equation of the tangent and normal lines for the following implicit function at the given point.

3 𝑥2+𝑦2=14

𝑦−√2=− 3√2 (𝑥− 2 )

(2 ,√2 )𝑑𝑦𝑑𝑥

=−3 𝑥𝑦

𝑚𝑡𝑎𝑛=𝑑𝑦𝑑𝑥

=−3 (2 )√2

𝑚𝑡𝑎𝑛=𝑑𝑦𝑑𝑥

=−6

√2=−3√2

𝑚𝑡𝑎𝑛=−6

√2=−3 √2

𝑚𝑛𝑜𝑟𝑚𝑎𝑙=√26

𝑦−√2=√26

(𝑥− 2 )


Find

𝑥35 𝑦4 𝑑𝑦𝑑𝑥

+¿

𝑥3 𝑦5+3 𝑥=8 𝑦3+1

3𝑑𝑥𝑑𝑥

=¿3 𝑥2 𝑑𝑥𝑑𝑥

𝑦5+¿ 24 𝑦2 𝑑𝑦𝑑𝑥

+0

5 𝑥3 𝑦4 𝑑𝑦𝑑𝑥

+3 𝑥2 𝑦5+3=24 𝑦2 𝑑𝑦𝑑𝑥

5 𝑥3 𝑦4 𝑑𝑦𝑑𝑥

−2 4 𝑦2 𝑑𝑦𝑑𝑥

=− 3 𝑥2 𝑦5 −3

𝑑𝑦𝑑𝑥

(5 𝑥3 𝑦4 −24 𝑦2 )=− 3𝑥2 𝑦5 −3

𝑑𝑦𝑑𝑥

= −3 𝑥2 𝑦5 −3 5 𝑥3 𝑦4 − 24 𝑦 2



The price (p in dollars) and the number of sales of a certain item (x) are related by the following equation, . Find the function that represents the rate of change of x with respect to time (t). Using that function, find dx/dt when x = 5, p = 8, and dp/dt = 2.6

5𝑝+4 𝑥+2𝑝𝑥=60

5𝑑𝑝𝑑𝑡

+4𝑑𝑥𝑑𝑡

+2𝑝𝑑𝑥𝑑𝑡

+2𝑑𝑝𝑑𝑡

𝑥=0

4𝑑𝑥𝑑𝑡

+2𝑝𝑑𝑥𝑑𝑡

=− 5𝑑𝑝𝑑𝑡

−2 𝑥𝑑𝑝𝑑𝑡

( 4+2𝑝 ) 𝑑𝑥𝑑𝑡

=(−5 − 2𝑥 ) 𝑑𝑝𝑑𝑡

𝑑𝑥𝑑𝑡

=(−5 −2 𝑥4+2𝑝 ) 𝑑𝑝𝑑𝑡

𝑑𝑥𝑑𝑡

=(−5 −2 (5 )4+2 (8 ) )2.6

𝑑𝑥𝑑𝑡

=− 1.95 𝑖𝑡𝑒𝑚𝑠/𝑑𝑎𝑦


1. Read the problem, pull out essential information and identify a formula to be used.

2. Sketch a diagram if possible.

3. Write down any known rate of change & the rate of change you are looking for. 4. Be careful with signs…if the amount is decreasing, the rate of change is negative. 5. Pay attention to whether quantities are constant or varying.

6. Set up an equation involving the appropriate quantities. 7. Differentiate with respect to t (could be other variables) using implicit differentiation. 8. Plug in known items (you may need to find some quantities using geometry). 9. Solve for the item you are looking for, most often this will be a rate of change. 10. State your final answer with the appropriate units.

Related rates problems involve finding a rate at which a quantity changes by relating that quantity to other quantities whose rates of change are known.

5) A balloon is being inflated at a rate of 10 cubic centimeters per second. How fast is the radius of a spherical balloon changing at the instant the radius is 5 centimeters?

𝑉=43𝜋 𝑟3

𝑑𝑉𝑑𝑡

=10 𝑐𝑚3/𝑠𝑒𝑐

𝐹𝑖𝑛𝑑𝑑𝑟𝑑𝑡

h𝑤 𝑒𝑛𝑟=5𝑐𝑚

𝑉=43𝜋 𝑟3

𝑑𝑉𝑑𝑡

=43𝜋 3𝑟2 𝑑𝑟

𝑑𝑡

10=4𝜋 (5 )2 𝑑𝑟𝑑𝑡

𝑑𝑟𝑑𝑡

=0.032𝑐𝑚 /𝑠𝑒𝑐

𝑑𝑉𝑑𝑡

=4𝜋 𝑟2 𝑑𝑟𝑑𝑡

10

4𝜋 (5 )2=𝑑𝑟𝑑𝑡



7) A 25-foot ladder is leaning against a wall. The bottom of the ladder is being pulled away from the wall at a rate of 3 feet per second. How fast is the top of the ladder moving at the instant the bottom of the ladder is 15 feet away from the wall?

𝑦

𝑥

𝑧

𝑥2+ 𝑦2=𝑧2

𝑧=25 𝑓𝑡𝑑𝑥𝑑𝑡

=3 𝑓𝑝𝑠

𝐹𝑖𝑛𝑑𝑑𝑦𝑑𝑡

h𝑤 𝑒𝑛𝑥=15 𝑓𝑡

𝑥2+ 𝑦2=𝑧2

𝑥2+ 𝑦2=252

2 𝑥𝑑𝑥𝑑𝑡

+2 𝑦𝑑𝑦𝑑𝑡

=0

2 (15 ) (3 )+2 𝑦𝑑𝑦𝑑𝑡

=0

152+ 𝑦2=252

𝑦=20

2 (15 ) (3 )+2 (20 ) 𝑑𝑦𝑑𝑡

=0

40𝑑𝑦𝑑𝑡

=− 90

𝑑𝑦𝑑𝑡

=−2.25𝑓 𝑝𝑠


1¿ 𝐼𝑓𝑥=𝑦3 −𝑦 𝑎𝑛𝑑𝑑𝑦𝑑𝑡

=5 , h𝑡 𝑒𝑛 h𝑤 𝑎𝑡 𝑖𝑠 h𝑡 𝑒𝑣𝑎𝑙𝑢𝑒𝑜𝑓𝑑𝑥𝑑𝑡

h𝑤 𝑒𝑛 𝑦=2 ?

𝑥=𝑦 3− 𝑦𝑑𝑥𝑑𝑡

=3 𝑦2 𝑑𝑦𝑑𝑡

−𝑑𝑦𝑑𝑡

𝑑𝑥𝑑𝑡

=3 (2 )2 (5 )−5

𝑑𝑥𝑑𝑡

=55

Example

Documents

Sec 2.6 – Marginals and Differentials 2012 Pearson Education, Inc. All rights reserved Let C(x), R(x), and P(x) represent, respectively, the total cost,