Section 4-1 The Constant e and Continuous Compound Interest

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Name Date Class Section 4-1 The Constant e and Continuous Compound Interest Goal: To work with the constant e and solve continuous compound problems Definition: The number e 1 e = lim 1+ n n n or e= lim ( 1+ s) 1 s 0 s Theorem 1: Continuous Compound Interest Formula If a principal P is invested at an annual rate r (expressed as a decimal) compounded continuously, then the amount A in the account at the end of t years is given by: A= Pe rt 1. Use a calculator to evaluate A to the nearest cent. 0.05t A= 1500e for t = 3, 6, 9 4-1

2. Solve for t to two decimal places. 4 = e 0.07t 3. Solve for r to two decimal places. 3 = e 4r 4. Bank A offers a 7 year CD that earns 3.75% compounded continuously. a.) If $5,000 is invested in this CD, how much will it be worth in 7 years? 4-2

b.) How long will it take for the account to be worth $10,000? 5. A note will be worth $25,000 at maturity 10 years from now. How much should you be willing to pay for the note now if money is worth 5.6% compounded continuously? 6. A woman invests $15,000 in an account that pays 7.8% compounded continuously. If the woman needs $20,000, how long will it take to accumulate that much money? 4-3

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Name Date Class Section 4-2 Derivatives of Exponential and Logarithmic Functions Goal: To find the first derivative of functions that are exponential or logarithmic in nature. Definition: Derivatives of Exponential and Logarithmic Functions For b> 0, b 1, d e dx x e x = d b x x = b ln b dx d 1 ln x = dx x d 1 1 logb x = dx ln b x 1-5 Find the first derivative for the given function. 1. x f( x) = 7e + 2x 5x+ 6 2 2. f( x) = 3e + 2ln x+ 7x 5 x 4-5

3. 3 2 5 f( x) = 4x + 2x 5x+ lnx + e x 4. x x+ 3 y = 10 e 5x 5. y = log x+ 2x 3 x 4 3 4-6

x 6. Given the function f( x) = 5e + 7, find the equation of the line tangent to the graph of the function at x = 0. 4-7

5 7. Given the function f( x) = lnx + 2, find the equation of the line tangent to the graph of the function at x= e. 4-8

Name Date Class Section 4-3 Derivatives of Products and Quotients Goal: To find the first derivative of functions that are written as products or quotients. Theorem 1: Product Rule If y = f( x) = F( x) S( x) and if F ( x) and S ( x) exist, then Theorem 2: Quotient Rule f ( x) = F( x) S ( x) + S( x) F ( x) If T( x) y = f( x) = and if T ( x) and B ( x) exist and Bx ( ) 0, then Bx ( ) f ( x) = BxT ( ) ( x) TxB ( ) ( x) [ Bx ( )] 2 1-10 Find the first derivative and simplify. 1. 3 2 f( x) = 3 x ( x 5x+ 3) 4-9

2. 2 2 f( x) = (4x + 7)(2x 9) 3. f( x) = 7x 2 x 6 4-10

4. 2x 5 f( x) = 2 3x + 5 5. f( x) = e ln x x 4-11

6. 3 2 2 y = (2x 5 x )( x 2 x) 7. y = x 2 3x+ 4 2 4x 3 4-12

8. f( x) = 8 log4 x x 9. f( x) = 8 2 4 x x + 4 4-13

10. 3 2 (2x 5)(3x + 1) f( x) = 2 x + 7 4-14

Name Date Class Section 4-4 The Chain Rule Goal: To find the first derivative of functions that are written as composite functions. Definition: Composite Function A function m is a composite of functions f and g if [ ] mx ( ) = f gx ( ) The domain of m is the set of all numbers x such that x is in the domain g and gxis ( ) in the domain of f. Theorem 1: General Power Rule If ( ) uxis a differentiable function, n is any real number, and y f x [ u x ] [ ] n 1 f ( x) = n u( x) u ( x) = ( ) = ( ), then n Theorem 2: Chain Rule If y = f( u) and u g( x) = define the composite function y m x f [ g x ] = ( ) = ( ), then dy dx dy du = provided dy du dx du and du dx exist 1. Find mx ( ) f[ gx ( )] = if f( u) 5 = u and 3 gx ( ) = 2x 6x+ 7. 4-15

2. Find mx ( ) f[ gx ( )] = if f( u) = lnuand 2 gx ( ) = x + 3x 5. 4 2 3 3. Given y = ( x + 3x 3x+ 5), write this composite function in the form y = f( u) and u = g( x). 4. Given u = g( x). 2 3x 2x 3, y = e + write this composite function in the form y = f( u) and 5 10 Find the first derivative for the given functions. 5. 3 4 f( x) = (2x 7x+ 3) 4-16

6. f( x) = 6ln(7x 9) 7. 2 1 f( x) = 6e x 4-17

8. 2 6 y = 6( x 3x+ 12) 9. 4 2 f( x) = x 2x 4-18

10. y = (ln(3x+ 7)) 3 4-19

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Name Date Class Section 4-5 Implicit Differentiation Goal: To find the first derivative of functions that are not explicit in nature. Procedure: 1. Find the derivative of the function using the basic derivative rules using y or dy for the derivative of the variable y. dx 2. Remove parentheses, if necessary 3. Group all terms with y on one side of the equation and group all terms without y on the other side of the equation. 4. Factor out y for all terms that have a y. 5. Solve for y by dividing. 1. Consider the equation 2 2x + 7y 9= 0 a.) Solve the equation for y and then find the first derivative. b.) Use implicit differentiation to find the first derivative. 4-21

2. Use implicit differentiation to find the first derivative and evaluate it at the given point. 3 2y 8x + 2= 0; (2, 1) 3. Use implicit differentiation to find the first derivative and evaluate it at the given point. 2 2 2xy + 3y 2x = 0; (1,4) 4-22

4. Use implicit differentiation to find the first derivative and evaluate it at the given point. x 2 2 5ln y = e + x y ; (0,3) 5. Use implicit differentiation to find the first derivative and evaluate it at the given point. 2 3 2 x y + xy 6y+ 4lnx 6= 0; ( 2,3) 4-23

6. Use implicit differentiation to find the first derivative and evaluate it at the given point. y 3 2 8+ 2x + 3= 0; (3,2) 7. Use implicit differentiation to find the first derivative and evaluate it at the given point. 4 3 (4 + y) + 6y = x 2x+ 7; (1,0) 4-24

Name Date Class Section 4-6 Related Rates Goal: To solve problems that involve related rates Procedure: Solving Related Rates Problems 1. Sketch a figure, if possible. 2. Identify all relevant variables, including those whose rates are given and those whose rates are to be found. 3. Express all given rates and rates to be found as derivatives. 4. Find an equation connecting the variables identified in step 2. 5. Implicitly differentiate the equation found in step 4, using the chain rule where appropriate, and substitute in all given values. 6. Solve for the derivative that will give the unknown rate. 1. Assume that x= x() t and y = y(). t Find dy dt given 2 3 dx x + 8xy y = 12, = 3, x= 1, y = 2. dt 4-25

2. A boat is being pulled toward a dock by a rope. If the rope is being pulled in at 2 feet per second and the rope is 3 feet above the dock, how fast is the distance between the dock and the boat decreasing when it is 25 feet from the dock? 4-26

3. A 20-foot ladder is placed against a vertical wall. Suppose that the top of the ladder slides down the wall at a constant rate of 4 feet per second. How fast is the bottom of the ladder sliding away from the wall when the top of the ladder is 12 feet above the ground? 4-27

4. The radius of a spherical balloon is increasing at a constant rate of 5 inches per minute. How fast is the volume changing when the radius is 15 inches? 4-28

Name Date Class Section 4-7 Elasticity of Demand Goal: To determine the elasticity of demand for functions Definition: Relative and Percentage Rates of Change The relative rate of change of a function f( x) is f ( x) The percentage rate of change is 100. f( x) f ( x). f( x) Theorem 1: Elasticity of Demand by: If price and demand are related by x= f( p), then the elasticity of demand is given E( p) = pf ( p) f( p) a. If 0 < E( p) < 1, then demand is inelastic. b. If E( p ) = 1, then demand is unit. c. If E( p ) > 1, then demand is elastic. 1. Given 2 f( x) = 33x 0.6 x, find the relative rate of change. 4-29

2 2. Given f( x) = 5x 6ln x, find the relative rate of change when x = 4. (Round to three decimals if necessary.) 3. Given f( x) = 600 7 x, find the relative rate of change when x = 60. (Round to three decimals if necessary.) 2 4. Given f( x) = 4500 3 x, find the percentage rate of change when x = 50. (Round to the nearest tenth if necessary.) 4-30

5. Given x= f( p) = 12,000 300 p, use the price demand equation to find E( p), the elasticity of demand. 2 6. Given x= f( p) = 850 3p 0.04 p, use the price demand equation to find E( p), the elasticity of demand to determine whether the demand is elastic, inelastic, or has unit elasticity at the indicated values of p. a. p = 15 b. p = 35 c. p = 55 4-31

2 7. Given x= f( p) = 2400 4 p, use the price demand equation to find the values of p for which demand is elastic and the values for which demand is inelastic. Assume that price and demand are both positive. 4-32

2 8. Given x= f( p) = 1600 2 p, use the price demand equation to find the values of p for which demand is elastic and the values for which demand is inelastic. Assume that price and demand are both positive. 4-33

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