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1 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.

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Presentation on theme: "1 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."— Presentation transcript:

1 1 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,

2 2 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

3 3 Examples. f (x) = 5 ln x. f (x) = x 5 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 )

4 4 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 OR We just learned that What if instead of x we had an ugly function?

5 5 Examples. f (x) = ln (x 4 + 5) f (x) = 4 ln √x f ‘ (x) = = 4 ln x 1/2

6 6 Examples. f (x) = (5 – 3 ln x) 4. = 4 (5 – 3 ln x) 3 f ‘ (x) = 4 (5 – 3 ln x) 3 f ‘ (x) =

7 7 Derivative Formulas for e x. The above derivative can be combined with the power rule, product rule, quotient rule, and chain rule to find more complicated derivatives

8 8 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 + e x 4x 3 Hint, use the product rule.

9 9 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 OR We just learned that What if instead of x we had an ugly function?

10 10 Example.

11 11 General Derivative Rules Power RuleGeneral Power Rule General Exponential Derivative RuleExponential Rule Log Rule General Log Derivative Rule

12 Maximizing Consumer Expenditure The 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.

13 13 Example Consumer 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

14 14 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

15 15 Summary. The derivative of f (x) = e x is f ' (x) = e x. The derivative of f (x) = e u is f ' (x) = e u 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.

16 16 ASSIGNMENT §3.5 on my website.


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