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Chapter 3 Additional Derivative Topics

Chapter 3 Additional Derivative Topics. Section 2 Derivatives of Exponential and Logarithmic Functions. The Derivative of e x. The process of finding the derivative of e x uses the fact that Using the four step derivative process Step 1: Find f ( x + h ) = e x+h = e x ·e h

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Chapter 3 Additional Derivative Topics

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  1. Chapter 3Additional DerivativeTopics Section 2 Derivatives of Exponential and Logarithmic Functions

  2. The Derivative of ex The process of finding the derivative of exuses the fact that Using the four step derivative process Step 1: Find f(x + h) = ex+h= ex·eh Step 2: Find f(x + h) – f(x) = ex·eh – ex = ex(eh – 1) Step 3: Step 4:

  3. The Derivative of ex The derivative of the exponential function, exis the exponential function ex. That is, Caution 1: When using technology (calculator, etc.) make sure that you use the designated e for the exponential function. Caution 2: The power rule cannot be used to differentiate the exponential function, ex, that is,

  4. Review of Properties of ln x Recall that the inverse of an exponential function is called a logarithmic function. For b > 0 and b ≠ 1, the logarithmic form y = logb x is equivalent to the exponential form x = by. The domain of the logarithmic function is (0, ∞) and the range is (–∞, ∞). The domain of the exponential function is (–∞, ∞) and the range is (0, ∞).

  5. Review of Properties of ln x The graphs of y = logb x and y = bxare symmetric with respect to the line y = x. y = bx y = x y = logb x Of all of the possible bases for logarithmic functions, the two most widely used are log x = log10x Common logarithm (base 10) ln x = loge x Natural logarithm (base e)

  6. Derivative of ln x Let f(x) = ln x, x > 0. Step 1: Find f(x + h) = ln (x + h) Step 2: Find f(x + h) – f(x) = ln (x + h) – ln (x) =

  7. Derivative of ln x Continued Step 3: (Multiply by 1 = x/x)

  8. Derivative of ln x Continued Step 4:

  9. Example Finding Derivatives (A) y = 3ex + 5 ln x Solution: (A) (B) y = x4 – ln x4 Solution: (B) Use a property of logarithms to rewrite. y = x4 – ln x4 = x4 – 4 ln x

  10. Other Logarithmic and Exponential Functions In most applications involving logarithmic and exponential functions, the number e is the preferred base. Some situations are better handled with a base other than e. Derivatives of y = logb x and y = bxare obtained by expressing these functions in terms of natural logarithmic and exponential functions.

  11. Change of Base for Logarithms Find a relationship between logb x and ln x for any base b > 0 and b ≠ 1. Write y = logbx in exponential form as by = x. ln by = ln x y ln b = ln x Therefore,

  12. Change of Base for Exponential Find a relationship between bxand exfor any base b such that b > 0, b ≠ 1. y = bx ln y = ln bx ln y = x ln b y = ex ln b

  13. Summary of Derivatives of Exponential and Logarithmic Functions

  14. Examples Find derivatives for (A) f (x) = log5x(B) g (x) = 2x – 3x Solution: (A) (B)

  15. Example Exponential Model An online store sells guitar strings. If the store sells x sets of guitar strings at a price of $p per set, then the price demand equation is p = 12.5(0.998)x. Find the rate of change of price with respect to demand when the demand is 100 sets of guitar strings. Interpret the result. Solution: When x = 100, The price per set of strings is decreasing at a rate of about 2 cents per set when the demand is 100 sets of strings.

  16. Example Continuous Compound Interest An investment of $10,000 earns interest at an annual rate of 4% compounded continuously. (A) Find the instantaneous rate of change of the amount in the account after 2 years. Solution: The amount A(t) at time t (in years) is given by A(t) = 10,000e0.04t A´(t) = 400e0.04·t A´(2) = 400e0.04·2 = 433.31 After two years, the account balance is growing at the rate of $433.31 per year.

  17. Example Continuous Compound Interest An investment of $10,000 earns interest at an annual rate of 4% compounded continuously. (B) Find the instantaneous rate of change of the amount in the account at the time the account has a balance of $20,000. Solution: Recall that the amount A(t) at time t (in years) is given by A(t) = 10,000e0.04tand A´(t) = 400e0.04·t = 0.04·10000· e0.04·t = 0.04·A(t), that is, for any time t, the rate of change in the amount is equal to 4% of the balance in the account. It follows that the rate of change in the account when the balance is $20,000 is 4% of $20,000 which is $800.

  18. Example Logarithm Model A model for the growth of a sandwich shop franchise is N(t) = –765 + 482 ln t where N(t) is the number of locations in year t (t = 0 corresponds to 1980). Estimate the number of locations in 2028 and interpret the result. Solution: The year 2028 corresponds to t = 48. N(48) = –765 + 482 ln 48 ≈ 1,101 There will be approximately 1,101 locations in the year 2028.

  19. Example Logarithm Model A model for the growth of a sandwich shop franchise is N(t) = –765 + 482 ln t where N(t) is the number of locations in year t (t = 0 corresponds to 1980). Estimate the rate of change in the number of locations in 2028. Interpret the result. Solution: The year 2028 corresponds to t = 48. N´(t) = N´(48) = The number of locations is growing at a rate of 10 per year in the year 2028.

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