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The student will learn about:

the derivative of ln x and the ln f (x),

applications.

§3.5 Derivatives of Logarithmic and Exponential Functions.

the derivative of ln x and the ln f (x), the derivative of e x and e f (x) and,

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Derivative Formula for ln x.

The above derivative can be combined with the power rule, product rule, quotient rule, and chain rule to find more complicated derivatives

x

1xln

dx

d

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Examples.

f (x) = 5 ln x.

f (x) = x5 ln x. Note: We need the product rule.

(x 5 )(1/x) + (ln x)(5x 4 )

f ‘ (x) = (5)(1/x) = 5/x

f ‘ (x) =

= x 4 + (ln x)(5x 4)

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Derivative Formula for ln f (x).

The above derivative can be combined with the power rule, product rule, quotient rule, and chain rule to find more complicated derivatives

d 1 d 1ln(u) u u'

dx u dx u OR

d 1ln(x)

dx xWe just learned that

What if instead of x we had an ugly function?

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Examples.

f (x) = ln (x 4 + 5)

f (x) = 4 ln √x

f ‘ (x) = )5x(dx

d

5x

1 44

f ‘ (x) = 431

x 54x

21x

14)x('f

x

2

= 4 ln x 1/2

5x

x44

3

xln2

1 21x

2

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Examples.

f (x) = (5 – 3 ln x) 4 .

= 4 (5 – 3 ln x) 3

f ‘ (x) = 4 (5 – 3 ln x) 3 )xln35(dx

d

3

x

f ‘ (x) =

x

)xln35(12 3

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Derivative Formulas for ex.

The above derivative can be combined with the power rule, product rule, quotient rule, and chain rule to find more complicated derivatives

xx eedx

d

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Examples. Find derivatives for

f (x) = 3 e x. f ‘ (x) = 3 e x .

f (x) = x 4 e x

f ‘ (x) = x 4 e x + ex 4x 3

Hint, use the product rule.

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Derivative Formulas for e f (x).

The above derivative can be combined with the power rule, product rule, quotient rule, and chain rule to find more complicated derivatives

u u ud de e u e u'

dx dx OR

x xde e

dxWe just learned that

What if instead of x we had an ugly function?

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Example.

)x('f

1x3x 23

e)x(f

3 2x 3x 1 2(3x 6xf '( e )x)

)1x3x(dx

de 231x3x 23

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General Derivative Rules

Power Rule General Power Rule

General Exponential Derivative Rule

1nn xnxdx

d 'uunudx

d 1nn

Exponential Rule

xx eedx

d 'uee

dx

d uu

Log Rule

x

1xln

dx

d

General Log Derivative Rule

'uu

1uln

dx

d

Maximizing Consumer ExpenditureThe amount of a commodity that consumers will buy depends on the price of the commodity.

For a commodity whose price is p, let the consumer demand be given by a function D(p). Multiplying the number of units D(p) by the price p gives the total consumer expenditure for the commodity.

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ExampleConsumer Demand and Expenditure.

The consumer expenditure, is E (p) = p · D (p), where D is the demand function.

Let consumer demand be

D (p) = 8000 e – 0.05 p

Graph this on your calculator and see if it makes sense.

0 ≤ x ≤ 15 and 0 ≤ y ≤ 6,000

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Consumer Demand and Expenditure. Continued

The consumer expenditure, is E (p) = p · D (p), where D is the demand function.

Let consumer demand be D (p) = 8000 e – 0.05 p

Maximize the consumer expenditure.

Consumer expenditure E (p) = p 8000 e – 0.05 p

Use your calculator to maximize this.

E (20) = $58,860.71 0 ≤ x ≤ 30 and 0 ≤ y ≤ 65,000

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Summary.

• The derivative of f (x) = ex is f ' (x) = ex.

• The derivative of f (x) = eu is f ' (x) = eu u'.

• The derivative of f (x) = ln x is f ' (x) = 1/x.

• The derivative of f (x) = ln u is f ' (x) = (1/u) u'.

• We did an application involving consumer expenditure.

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ASSIGNMENT

§3.5 on my website.