o gr. g or. volkan - deukisi.deu.edu.tr/volkan.oger/fba1201/chapter11_slayt... · 2014-12-03 · dc...

Ogr. Gor. Volkan OGER FBA 1021 Calculus 1/ 55

Figure 1: x→ af(x)− f(a)

x− a=?


If the difference x− a is called h, then we can write x as a + h.Here we must have h 6= 0, for if h = 0, then x = a. We can writein the following form

x→ af(x)− f(a)

x− a=

f(a + h)− f(a)

hh→ 0


Definition

The derivative of a function f is the function denoted f ′(read ”fprime”) and defined by

f ′(x) = limh→0

f(x + h)− f(x)

h(1)

provided that this limit exists.

If f ′(a) can be found, f is said to be differentiable at a, and f ′(a)is called the derivative of f at a.


Besides the notation f ′(x), other common ways to denote thederivative of y = f(x) at x are

dy

dxpronounced ”dee y, dee x” or ”’dee y by dee x”

d

dx(f(x)) ”dee fx), dee x” or ”dee by dee x of f(x)”

y′ ”y prime”

Dxy ”dee x of y”

Dx(f(x)) ”dee x of f(x)”


Definition

f ′(a) is the slope of the line tangent to the graph of y = f(x) at( a , f(a) )


Rules for Differentiation

If c is a constant, thend

dx(c) = 0

That is, the derivative of a constant function is zero.


Example

d

dx(3) = 0 because 3 is a constant function.

Example

if g(x) =√

5, then g′(x) = 0 because g is a constant function.


If a is any real number, then

d

dx(xa) = a xa−1

That is, the derivative of a constant power of x is the exponenttimes x raised to a power one less than the given power.


Example

d

dx(x2) = 2 x2−1 = 2x.

Example

if g(x) = x = x1, then g′(x) = 1 x1−1 = 1 x0 = 1 Thus, thederivative of x with respect to x is 1.

Example

if f(x) = x−10, then f ′(x) = −10 x−10−1 = −10x−11


Example

if y =√x, then

y′ =d

dx(√x) =

d

dx(x

12 ) =

1

2x

12−1 =

1

2x−

12 =

1

2√x


Example

if h(x) =1

x√x

, then

h′(x) =d

dx(x

−32 ) =

−3

2x

−32−1 =

−3

2x−

52


if f is a differentiable function and c is a constant, then cf(x) isdifferentiable, and

d

dx(cf(x)) = cf ′(x)

That is, the derivative of a constant times a function is theconstant times the derivative of the function.


Example

if g(x) = 5x3, then

g′(x) =d

dx(5x3) = 5

d

dx(x3) = 5(3x3−1) = 15x2


Example

if y =0.255√x2

, then

y = 0.251

5√x2

= 0.25 x−2/5

y′ =d

dx(0.25x−2/5) = 0.25

d

dx(x−2/5)

= 0.25(−2

5x

−25−1)

= −0.1x−75


if f and g are differentiable functions, then f + g and f − g aredifferentiable, and

d

dx(f(x)∓ g(x)) = f ′(x)∓ g′(x)

That is, the derivative of the sum (difference) of two functions isthe sum (difference) of their derivatives.


Example

if F (x) = 3x5 +√x, then

F ′(x) =d

dx(3x5) +

d

dx(x1/2)

= 3d

dx(x5) +

d

dx(x1/2)

= 3(5x4) +1

2(x−1/2)

= 15x4 +1

2√x


Example

if f(z) =z4

4− 5

z1/3, then

f ′(z) =d

dz(1

4z4)− d

dz(5z

−13 )

=1

4

d

dz(z4)− 5

d

dz(z

−13 )

=1

4(4 z3)− 5(

−1

3z

−13−1)

= z3 − −5

3z

−43 = z3 +

5

3z

−43


Example

Find an equation of the tangent line to the curve

y =3x2 − 2

x

when x = 1.

First we find the derivative of

y =3x2 − 2

x=

3x2

x− 2

x= 3x− 2x−1.

dy

dx=

d

dx

(3x− 2x−1

)=

d

dx(3x)− d

dx(2x−1)

= 3d

dx(x)− 2

d

dx(x−1) = 3(1) + 2(x−2)


The slope of the tangent line to the curve when x = 1 is

dy

dx

∣∣∣x=1

= 3 +2

12= 5

To find y coordinate of the point on the curve where x = 1, weevaluate y = 3x2−2

x at x = 1. This gives

y =3(1)2 − 2

1= 1

Hence, the point (1, 1) lies on both the curve and the tangent line.Therefore, an equation of the tangent line is

y − 1 = 5(x− 1)


Applications of Differentiation to Economics

A manufacturer’s total-cost function, c = f(q), gives the total costc of producing and marketing q units of a product. The derivativeof c with respect to q is called the marginal cost Thus,

marginal cost =dc

dq


For example, suppose c = f(q) = 0.1q2 + 3 is a cost function,where c is in dollars and q is in pounds. Then

dc

dq= 0.2q

The marginal cost when 4 lb are produced is dc/dq, evaluatedwhen q = 4:

dc

dq

∣∣∣q=4

= 0.2(4) = 0.80


This means that if production is increased by 1 lb, from 4 lb to 5lb, then the change in cost is approximately $0.80. That is, theadditional pound costs about $0.80. In general, we interpretmarginal cost as the approximate cost of one additional unit ofoutput.


Example

If a manufacturer’s average-cost equation is

c = 0.0001q2 − 0.02q + 5 +500

q

find the marginal-cost function. What is the marginal cost when50 units are produced?


The marginal-cost function is the derivative of the total-costfunction c. Thus, we first find c by multiplying c by q. We have

c = q × c = q

(0.0001q2 − 0.02q + 5 +

500

q

)c = 0.0001q3 − 0.02q2 + 5q + 500


Differentiating c, we have the marginal-cost function:

dc

dq= 0.0001(3q2)− 0.02(2q) + 5 = 0.0003q2 − 0.04q + 5

The marginal cost when 50 units are produced is

dc

dq

∣∣∣q=50

= 0.0003(50)2 − 0.04(50) + 5 = 3.75

If c is in dollars and production is increased by one unit, fromq = 50 to q = 51,then the cost of the additional unit isapproximately $3.75. If production is increased by 1

3 unit, fromq = 50, then the cost of the additional output is approximately

(13)(3.75) = $1.25.


Suppose r = f(q) is the total-revenue function for a manufacturer.The equation r = f(q) states that the total dollar value receivedfor selling q units of a product is r. The marginal revenue isdefined as the derivative of the total dollar value received withrespect to the total number of units sold.

marginal revenue =dr

dq

Marginal revenue indicates the rate at which revenue changes withrespect to units sold. We interpret it as the approximate revenuereceived from selling one additional unit of output.


Example

Suppose a manufacturer sells a product at $2 per unit. If q unitsare sold, the total revenue is given by

r = 2q

The marginal-revenue function is

dr

dq=

d

dq(2q) = 2

which is a constant function. Thus, the marginal revenue is 2regardless of the number of units sold. This is what we wouldexpect, because the manufacturer receives $2 for each unit sold.


The Product Rule and the Quotinet Rule

If f and g are differentiable functions, then the product f · g isdifferentiable, and

d

dx(f(x) · g(x)) = f ′(x) · g(x) + f(x) · g′(x)

That is, the derivative of the product of two functions is thederivative of the first function times the second, plus the firstfunction times the derivative of the second:

d

dx(product) =

(derivative

of first

)(second) + (first)

(derivativeof second

)


Example

If F (x) = (x2 + 3x)(4x + 5), find F ′(x)

We will consider

F (x) = (x2 + 3x)︸︷︷︸f(x)

(4x + 5)︸︷︷︸g(x)

Therefore, we can apply the product rule:

F ′(x) = f ′(x)g(x) + f(x)g′(x)

=d

dx(x2 + 3x)︸︷︷︸

derivativeof first

(4x + 5)︸︷︷︸second

+ (x2 + 3x)︸︷︷︸first

d

dx(4x + 5)︸︷︷︸

derivativeof second

= (2x + 3)(4x + 5) + (x2 + 3x)(4) = 12x2 + 34x + 15


Example

If (x2/3 + 3)(x−1/3 + 5x), find dy/dx

dy

dx=

d

dx(x2/3 + 3)(x−1/3 + 5x) + (x2/3 + 3)

d

dx(x−1/3 + 5x)

=

(2

3x−1/3

)(x−1/3 + 5x) + (x2/3 + 3)

(−1

3x−4/3 + 5

)

=25

3x2/3 +

1

3x−2/3 − x−4/3 + 15


Example

If y = (x + 2)(x + 3)(x + 4), find y′

We would like to use the product rule, but as given it applies onlyto two factors. By treating the first two factors as a single factor,we can consider y to be a product of two functions:

y =(

(x + 2)(x + 3)︸︷︷︸f(x)

)(x + 4)︸︷︷︸

g(x)


The product rule gives

y′ =d

dx

((x + 2)(x + 3)

)(x + 4) +

((x + 2)(x + 3)

) d

dx(x + 4)

=d

dx

((x + 2)(x + 3)

)(x + 4) +

((x + 2)(x + 3)

)(1)

Applying the product rule again, we have

y′ =( d

dx(x + 2) · (x + 3) + (x + 2) · d

dx(x + 3)

)(x + 4)

+(x + 2)(x + 3)

=(

(1) · (x + 3) + (x + 2) · (1))

(x + 4) + (x + 2)(x + 3)

= 3x2 + 18x + 26


If f and g are differentiable functions and g(x) 6= 0, then thequotient f/g is also differentiable, and

d

dx

(f(x)

g(x)

)=

f ′(x)g(x)− f(x)g′(x)

(g(x)2)


Example

If F (x) =4x2 + 3

2x− 1, Find F ′(x)

Let f(x) = 4x2 + 3 and g(x) = 2x− 1. Then apply the quotientrule

F ′(x) =f ′(x)g(x)− f(x)g′(x)

(g(x)2)

=

d

dx(4x2 + 3) · (2x− 1)− (4x2 + 3) · d

dx(2x− 1)

(2x− 1)2

=(8x) · (2x− 1)− (4x2 + 3) · (2)

(2x− 1)2=

8x2 − 8x− 6

(2x− 1)2


Example

If the demand function for a manufacturer’s product is

p =1000

q + 5

where p is in dollars, find the marginal-revenue and evaluate itwhen q = 45.

The revenue function is

r =

1000

q + 5︸︷︷︸price

· q︸︷︷︸quantity

=1000q

q + 5


Thus, the marginal-revenue function is given by

dr

dq=

(q + 5) ddq (1000q)− (1000q) d

dq (q + 5)

(q + 5)2

dr

dq=

(q + 5)(1000)− (1000q)(1)

(q + 5)2=

5000

(q + 5)2

anddr

dq

∣∣∣q=45

=5000

(45 + 5)2=

5000

2500= 2

This means that selling one additional unit beyond 45 results inapproximately $2 more in revenue.


The Chain Rule

Our next rule, the chain rule, is ultimately the most important rulefor finding derivatives. It involves a situation in which y is afunction of the variable u, but u is a function of x and we want tofind the derivative of y with respect to x. For example, theequations

y = u2, and u = 2x + 1

define y as a function of u and u as a function of x. If wesubstitute 2x + 1 first equation, we can consider y to be a functionof x:

y = (2x + 1)2


To find dy/dx, we first expand (2x + 1)2:

y = 4x2 + 4x + 1

thendy

dx= 8x + 4

From this example, you can see that finding dy/dx by firstperforming a substitution could be quite involved. For instance, iforiginally we had been given y = x100 instead of y = u2, wewouldn’t even want to try substituting. Fortunately, the chain rulewill allow us to handle such situations with ease.


The Chain Rule

If y is a differentiable function of u and u is a differentiablefunction of x, then y is a differentiable function of x and

dy

dx=

dy

du· dudx


We will now use the chain rule to redo previous example. If

y = u2, and u = 2x + 1

thendy

dx=

dy

du· dudx

=d

du(u2)︸︷︷︸2u

· ddx

(2x + 1)︸︷︷︸2

= 4u

Replacing u by 2x + 1 gives

dy

dx= 4(2x + 1) = 8x + 4

which agrees with our previous result.


Example

If y =√w and w = 7− t3, find dy/dt

Solution: Here, y is a function of w and w is a function of t, so wecan view y as a function of t. By the chain rule,

dy

dt=

dy

dw· dwdt

=d

dw(√w) · d

dt(7− t3)

=

(1

2w−1/2

)(−3t2) =

1

2√w

(−3t2)

= − 3t2

2√

7− t3


Example

If y = (x3 − x2 + 6)100, then find y′

we think of the function as a composition. Let

y = f(u) = u100 and u = g(x) = x3 − x2 + 6

Then y = (x3 − x2 + 6)100 = (g(x))100 = f(g(x)). By the chainrule we have

dy

dx=

dy

du· dudx

= (100u99)(3x2 − 2x)

= 100(x3 − x2 + 6)99(3x2 − 2x)


Example

The cost c of producing q units of a product is given by

c = 5500 + 12q + 0.2q2

If the price per unit p is given by the equation

q = 900− 1.5p

use the chain rule to find derivative of cost with respect to priceper unit when p = 85.


Example

Suppose pq = 100 is the demand equation for a manufacturer’sproduct. Let c be the total cost, and assume that the marginalcost is 0.01 when q = 200. Use the chain rule to find dc/dp whenq = 200.


Derivatives of Logarithmic Functions

Derivative of ln(x)

d

dx(ln(x)) =

1

x, for x > 0


Example

Differentiate y =ln(x)

x2

By the quotient rule,

y′ =x2

d

dx(ln(x))− (ln(x))

d

dx(x2)

(x2)2

=

x2(

1

x

)− (ln(x))(2x)

x4

=x− 2x ln(x)

x4=

1− 2 ln(x)

x3for x > 0


Example

Differentiate y = ln(x2 + 1)

This function has the form y = ln(u) with u = x2 + 1, and sincex2 + 1 > 0, for all x, y = ln(x2 + 1) is defined. Using Chain Rule

dy

dx=

dy

du· dudx

dy

dx=

d

du(ln(u)) · d

dx(x2 + 1)

dy

dx=

1

u· (2x) =

1

x2 + 1· (2x) =

2x

x2 + 1


Derivative of Exponential Functions

d

dx(ex) = ex


Example

If y =x

ex, find

dy

dx

dy

dx=

exd

dx(x)− x

d

dx(ex)

(ex)2

dy

dx=

ex(1)− x(ex)

(e2x)

dy

dx=

ex(1− x)

(e2x)=

(1− x)

(ex)


Example

Findd

dx

(ex

3+3x)

The function has the form y = eu with u = x3 + 3x. By chain rule

dy

dx=

dy

du· dudx

= eu · (3x2 + 3) = ex3+3x(3x2 + 3)


Implicit Differentation

Implicit differentiation is a technique for differentiating functionsthat are not given in the usual form y = f(x) [nor in the formx = g(y)].An equation of the form F (x, y) = 0, such as we had originally, issaid to express y implicitly as a function of x. The word implicitly isused, since y is not given explicitly as a function of x. For example

x2 + y2 − 4 = 0


For F (x, y) = 0, the derivative of y with respect to x can beevaluated by

dy

dx= −

dFdxdFdy


Example

Finddy

dxby implicit differentiation if y + y3 − x = 7

Here F (x, y) = y + y3 − x− 7. Then

dF

dx= −1

dF

dy= 1 + 3y2

Thereforedy

dx= −

dFdxdFdy

= − −1

1 + 3y2=

1

1 + 3y2


Example

Find the slope of the curve x3 = (y − x2)2 at the point (1, 2)

Here F (x, y) = x3 − (y − x2)2. Then

dF

dx= 3x2 + 4x(y − x2)

dF

dy= −2(y − x2)

Therefore

dy

dx= −

dFdxdFdy

= −3x2 + 4x(y − x2)

−2(y − x2)=

x(3x + 4(y − x2))

2(y − x2)

The slope is

dy

dx

∣∣∣x=1,y=2

=(1)(3(1) + 4((2)− (12)))

2((2)− (12))=

7

2


o gr. g or. volkan - deukisi.deu.edu.tr/volkan.oger/fba1201/chapter11_slayt... · 2014-12-03 · dc...

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