Unit 4: Rules of Differentiation
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1 Unit : Rules of Differentiation DAY TOPIC ASSIGNMENT Power Rule p. Power Rule Again p. Even More Power Rule p. 5 QUIZ 5 Rates of Change p Rates of Change p QUIZ 8 Product Rule p. 0-9 Quotient Rule p. - 0 Both Product and Quotient Rules p. QUIZ Chain Rule p. 5 More Chain Rule p. 6 Still More Links p QUIZ 6 Higher Order Derivatives p. - 7 Even Higher Derivatives p. - 8 Review Worksheet (Passed out in class) 9 TEST
2 . Techniques of Differentiation Learning Objectives A student will be able to: Use various techniques of differentiations to find the derivatives of various functions. Compute derivatives of higher orders. Up to now, we have been calculating derivatives by using the definition. In this section, we will develop formulas and theorems that will calculate derivatives in more efficient and quick ways. It is highly recommended that you become very familiar with all of these techniques. The Derivative of a Constant If where is a constant, then. In other words, the derivative or slope of any constant function is zero. Proof: Eample : If for all, then for all. We can also write. The Power Rule If is a positive integer, then for all real values of The proof of the power rule is omitted in this tet, but it is available at and also in video form at Khan Academy Proof of the Power Rule. Note that this proof depends on using the binomial theorem from Precalculus.
3 . Eample : If, then and The Power Rule and a Constant If is a constant and is differentiable at all, then In simpler notation, In other words, the derivative of a constant times a function is equal to the constant times the derivative of the function. Eample : Eample :
4 Derivatives of Sums and Differences If and are two differentiable functions at, then and In simpler notation, The Product Rule If and are differentiable at, then In a simpler notation, The derivative of the product of two functions is equal to the first times the derivative of the second plus the second times the derivative of the first. Keep in mind that Eample 7:
5 Find for Solution: There are two methods to solve this problem. One is to multiply the product and then use the derivative of the sum rule. The second is to directly use the product rule. Either rule will produce the same answer. We begin with the sum rule. Taking the derivative of the sum yields Now we use the product rule, which is the same answer. The Quotient Rule If and are differentiable functions at and, then In simpler notation, The derivative of a quotient of two functions is the bottom times the derivative of the top minus the top times the derivative of the bottom all over the bottom squared. Keep in mind that the order of operations is important (because of the minus sign in the numerator) and Eample 8: Find for 5
6 Solution: Eample 9: At which point(s) does the graph of have a horizontal tangent line? Solution: Since the slope of a horizontal line is zero, and since the derivative of a function signifies the slope of the tangent line, then taking the derivative and equating it to zero will enable us to find the points at which the slope of the tangent line equals to zero, i.e., the locations of the horizontal tangents. Multiplying by the denominator and solving for, Therefore the tangent line is horizontal at Higher Derivatives If the derivative of the function is differentiable, then the derivative of, denoted by, is called the second derivative of. We can continue the process of differentiating derivatives and obtain third, fourth, fifth and higher derivatives of. They are denoted by,,,, Eample 0: Find the fifth derivative of. 6
7 Solution: Eample : Show that satisfies the differential equation Solution: We need to obtain the first, second, and third derivatives and substitute them into the differential equation. Substituting, which satisfies the equation. Review Questions Use the results of this section to find the derivatives... y =.. (where a and b are constants) Newton s Law of Universal Gravitation states that the gravitational force between two masses (say, the earth and the moon), m and M, is equal to their product divided by the square of the distance r between them. Mathematically, where G is the Universal Gravitational Constant. If the distance r between the two masses is changing, find a formula for the instantaneous rate of change of F with respect to the separation distance r. 7
8 . Find, where is a constant.. Find y (), where Review Answers y. (some answers simplify further than the given responses)
9 Constant and Power Rule Practice Use the Constant and Power Rules to find the derivative.. y. f ( ). f ( t) t t. s t t t ( ) 5. y t 6. f ( ) 7. y 8. y 9. y ( ) y. f ( ). f ( ). f( ). y 5. f ( ) f ( ) Find the value of the derivative of the function at the indicated point. 7. ( ) f, 8. f( t), t Find the equation of the tangent line to the graph of the function at the indicated point 9. y (,8) 0. y( ) (0,) Answers: ' 6. y 0. f '( ). f '( t) 6t. s'( t) t 5. y' t 8 6. f '( ) 7. y' 8. y ' 9. y ' 0. y ' f '( ). f '( ). f '( ). y' 5 5. f '( ) 6. f '( ) y y 9. 9
10 Power Rule Practice Find the derivative of each function. In your answers, rational eponents are OK, negative eponents are not..) f( ).) f ( ).) f( ).) f( ) POWER RULE: d a n na n, n, a d 5.) f( ) 6.) f( ) ( ) 7.) f( ) 9 7 Answers:.) f( ) 0.) f( ).) f( ).) f( ) 5.) f ( ) 6.) f( ) ) f( )
11 Mo Power Rule Practice Find the derivative of each function. Make sure your answers are factored completely. If a point is given, find the value of the derivative at that point..) f( ) Point:,.).).) 5.) 6.) 7.) 8.) f ( ) ( ) Point: (0, ) Point: (, ) f ( ) f( ) f ( ) ( ) Point: (7, 50) f ( ) ( )( ) f ( ) ( ) f( ) Point: (, 5) Answers:.) f ( ), f 6.) f ( ) 8 ( ), f (0) 8.) f ( ), f () ) 5.) 6.) 7.) 8.) 6 ( ) f( ) f f ( ), (7) 8 f ( ) 6 f f ( ) 9( ), ( ) 576 f( )
12 Product Rule Practice Find the derivative using the Product Rule. Final answer should be in simplest form.. f ( ) ( )( 5). f ( ). f ( ) 5. f ( ) 5 Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. 5. y 6. y Find the derivative. Do not use the Product Rule. 7. f ( ) 8. y ( ) Answers:. f '( ) f '( ). f '( ) f '( ) ,,,,, 6. none 7. f '( ) dy 8. 5 d
13 Product Rule Practice Find the derivative of each function. Make sure your answers are factored completely. If a point is given, find the value of the derivative at that point..).).).) 5.) 6.) f ( ) ( ) Point: (, ) f Point: (0, ) ( ) ( ) f ( ) ( )( ) Point: (, 6) f ( ) ( )( 5) f 5 ( ) ( )( 7 ) PRODUCT RULE: f ( ) ( ) Point: (, 6) d f ( ) g ( ) f ( ) g ( ) g ( ) f ( ) d 7.) f ( ) 8.) f ( ) ( )( 5)( ) Answers:.).).).) 5.) 6.) f f ( ) (5 ), () f f ( ), (0) 0 f f ( ), () f ( ) f 6 5 ( ) f f ( ) 6 ( ), () 5 7.) f( ) ) f ( ) 5 8 0
14 Quotient Rule Practice Use the Quotient Rule to find the derivative. Final answers should be in simplest form.. f( ). f( ). f ( ) Find the equation of the line tangent to f( ) at the indicated point.. f( ) 5 Point: (6, 6) 5. 5 f( ) Point: (, 6) Find the derivative without the use of the Product or Quotient Rules. Give simplified final answers f ( ) ( ) 7. f( ) Answers:. f '( ) 5. f '( ) y y 7. f '( ) 8 7. f '( ) f '( )
15 Quotient Rule Practice Find the derivative of each function. Make sure your answers are factored completely. If a point is given, find the value of the derivative at that point..) f( ) Point: (6, 6) 5.) 5 f( ) Point: (, 6).) f( ) QUOTIENT RULE: d f ( ) g( ) f ( ) f ( ) g( ) d g( ) g ( ).) f( ) 5.) 5 f ( ) ( ) (Do not use the product or quotient rules.) 6.) 7.) f( ) (Do not use the quotient rule.) f( ) Point: (, ) 8.) f ( ) Answers: 5.) f ( ), f (6) 5 ( 5) 5.) f( ) ( ).) f ( ), f () 6.) f( ) 5.) f( ) ( ) 7.) f ( ), f () ( ).) f( ) ( ) 8.) f( ) 8 7 ( ) 5
16 Practice Problems (Constant, Power, Product & Quotient Rules) Differentiate. Remember to simplify the function to make differentiating easier. Final answers should be in simplest form.. y. y. f ( ). y ( ) 5. y 6. y 7. f( ) 8. y 6 t 9. ht () 0. f( ) t 5t6. s() t. f ( ) ( )( ) Find the equation of the line tangent to the function at the indicated value.. f( ) Find the slope of the graph at the indicated point.. y ( )( ), Find the point(s), if any, at which the graph of the function has a horizontal tangent. 5. f( ) Answers:. y' 0. y' 6. f '( ). y ' 9 dy 5. y ' f '( ) 8. y ' d h'( t) 0. f '( ). f '( ) ( t ) ( ) 5 8. s'( t). y. m ( ) 9 0,,,,, 5. 6
17 . The Chain Rule Learning Objectives A student will be able to: Know the chain rule and its proof. Apply the chain rule to the calculation of the derivative of a variety of composite functions. We want to derive a rule for the derivative of a composite function of the form in terms of the derivatives of f and g. This rule allows us to differentiate complicated functions in terms of known derivatives of simpler functions. The Chain Rule If is a differentiable function at and is differentiable at, then the composition function is differentiable at. The derivative of the composite function is: Another way of epressing, if and, then And a final way of epressing the chain rule is the easiest form to remember: If is a function of and is a function of, then Eample : Differentiate Solution: Using the chain rule, let Then The eample above is one of the most common types of composite functions. It is a power function of the type The rule for differentiating such functions is called the General Power Rule. It is a special case of the Chain Rule. 7
18 The General Power Rule if then In simpler form, if then Eample : What is the slope of the tangent line to the function that passes through point? Solution: We can write This eample illustrates the point that can be any real number including fractions. Using the General Power Rule, To find the slope of the tangent line, we simply substitute into the derivative: Eample : Find for. Solution: The function can be written as Thus 8
19 Eample : Find for Solution: Let By the chain rule, where Thus Eample 5: Find for Solution: This eample applies the chain rule twice because there are several functions embedded within each other. Let be the inner function and be the innermost function. Using the chain rule, Notice that we used the General Power Rule and, in the last step, we took the derivative of the argument. 9
20 Multimedia Links For an introduction to the Chain Rule (5.0), see Khan Academy, Calculus: Derivatives : The Chain Rule (9:). For more eamples of the Chain Rule (5.0), see [ Math Video Tutorials by James Sousa The Chain Rule: Part of (8:5). and [ Math Video Tutorials by James Sousa The Chain Rule: Part of (8:5). Review Questions Find f ( ) sin 0
21 Review Answers.. f( ) ( sin cos ) sin f( ) or f( ) sin sin cos( ) cos sin( )sin f( ) cos ( ) Or f ( ) f( ) 5
22 Chain Rule Practice Differentiate. Simplify answers.. f ( ) 9. f ( ). 6 st () t 5. y. g ( ) 6. f ( ) 7. y csc 8. sin y cos 9. y 0. y tan 6 5. Find the equation of the tangent line to the graph of f( ). Find the slope of y sec when. 6 Answers:. f '( ) s'( t) 7 t y '.. f '( ) g'( ) 7 6. f '( ) ( )(5 ) csc cot 7. y ' 8. y' sin cos 9. y' sin cos 0. y' sec. 6 8 y at the point, () f.
23 Day Find the derivative of each function. ) y ) f( ) ) f ( ) ) g( ) 5) g ( ) 6) y t t 7) f ( t) t t 8) y 9 9) s t t t ( ) 0) y ) y t ) y 5 ) f ( ) ) g( ) 5) y 6) y 7) y 8) y
24 Day Find f (). ) f ( ) ) f ( ) 5 ) ) f ( ) f ( ) 5) f( ) 6) f( ) 7) f ( ) ( ) 8) f ( ) ( )( ) 9) f ( ) 5 5 0) f ( ) Find the value of the derivative at the indicated point. 7 ) f( ), (,) ) f ( ), (0, ) 5 ) f ( ), (,8) ) f ( ) ( ), (0,)
25 Day Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. ) y ) y 5 ) y ) y 5) The variable cost for manufacturing an electrical component is $7.75 per unit, and the fied cost is $500. Write the cost C as a function of, the number of units produced. Show that the derivative of this cost function is constant and is equal to the variable cost. (This is called the marginal cost. A marginal cost is the derivative of the cost) 6) A college club raises funds by selling candy bars for $.00 each. The club pays $0.60 for each candy bar and has annual fied costs of $50. Write the profit P as a function of, the number of candy bars sold. Show that the derivative of the profit function is a constant and that it is equal to the profit on each candy bar sold. (This is called the marginal profit. A marginal profit is the derivative of the profit) Using what you have learned about derivatives as well as your calculator. Find any points on the given interval of for which the function f has a horizontal tangent line. 7) f ( )..5, [0,] 8) f ( ) , [, ] 5
26 Day 5 ) The effectiveness E (on a scale from 0 to ) of a pain-killing drug t hours after entering the bloodstream is given by E (9 t t t ), 0 t.5. Find the average rate of change E on the intervals [, ] and 7 compare this with the instantaneous rates of change at the endpoints of the interval. ) At 0 Celsius, the heat loss H (in kilocalories per square meter per hour) from a person s body can be modeled by H (0 v v 0.5), where v is the wind speed (in meters per second). Find the instantaneous rate of change H when v = and when v = 5. ) The height s (in feet) at a time t (in seconds) of a silver dollar dropped from the top of the Washington Monument is s 6t 555. a) Find the average velocity on the interval [,]. b) Find the instantaneous velocity when t = and when t =. c) How long will it take the dollar to hit the ground? d) Find the velocity of the dollar when it hits the ground. *CONTINUE ON NEXT PAGE 6
27 ) The height s (in feet) of an object fired straight up from the ground level with an initial velocity of 00 feet per second is given by s 6t 00t, where t is the time (in seconds). a) How fast is the object moving after one second? b) During which interval of time is the speed decreasing? c) During which interval of time is the speed increasing? 5) The position s (in feet) of an accelerating car is s 0 t, 0 t 0, where t is the time (in seconds). Find the velocity of the car at the following times. a) t = 0 b) t = c) t = d) t = 9 6) Given the position of an object is given by s( t) t t find the equation for the velocity of the object. 7
28 Day 6 ) Given the cost function C( ) 55, , find the marginal cost function. ) Given the revenue function R( ) , find the marginal revenue function. ) Given the profit function P( ) ,000, find the marginal profit function. ) The revenue (in dollars) from producing units of a product is R( ) a) Find the additional revenue when production is increased from 5,000 units to 5,00 units. b) Find the marginal revenue when = 5,000. c) Compare your results from parts a & b. 5) The cost (in dollars) of producing units of a product is C( ) a) Find the additional cost when the production increases from nine to ten units. b) Find the marginal cost when = 9. c) Compare your results from parts a & b. *CONTINUE ON NEXT PAGE 8
29 6) The profit (in dollars) from selling units of a product is P( ) a) Find the additional profit when the sales increase from seven to eight units. b) Find the marginal revenue when = 7. c) Compare your results from parts a & b. 7) The profit (in dollars) for selling units of a product is given by P( ) Find the marginal profit for the following sales. a) = 0 b) = 0 c) = d) = 5 9
30 Day 8 Differentiate each function. ) f ( ) ( )( 5) ) f 5 ( ) ( )( 7 ) ) f ( ) ( ) ) f ( ) ( )( ) 5) f ( ) ( ) 6) 5 f ( ) ( ) Find the value of the derivative of the function at the indicated point. f ( ) at(,) f ( ) 5 at(,6) 7) 8) 9) f ( ) at(,0) *CONTINUE ON NEXT PAGE 0
31 0) Find the equation of the line tangent to f ( ) at (0,) ) Find the equation of the line tangent to f ( ) at (, -)
32 Day 9 Differentiate each function. ) f( ) ) f( ) ) f( ) ) f ( ) *Hint: multiply first then use Quo.Rule 5) f( ) 6) f( ) 5-You don t have to use Quo.Rule if you don t want. You can distribute instead if you like. Or you can use Quo.Rule too if you like. *CONTINUE ON NEXT PAGE
33 Find the value of the derivative of the function at the indicated point. 7) f ( ) at(6,6) 8) f ( ) at(, ) 5 9) 5 f ( ) at(, ) 6 0) Find the equation of the line tangent to f( ) at (,) ) Find the equation of the line tangent to f( ) at (,/)
34 Day 0 Find the derivative of each function. ) y ) y ) f ( ) Hint: You will need to use both! Or you will Have to multiply carefully then use the Quo.Rule. ) y Find the point(s), if any, at which the graph of f has a horizontal tangent. 5) f( ) 6) f( ) 7) The percent P of defective parts produced by a new employee t days after the employee starts work can be t 750 modeled by P. Find the rate of change P when t = 0. 50t 00
35 Day Find the derivative of each function. ) y 7 ) y y ) y 6 ) y 9 6) y 5) 7) y 8) y 5 9) y 0) y 5 ) Find the equation of the line tangent to f ( ) 7 at =. Find the derivative of each function. (Hint: Rewrite them so they are in terms of negative powers instead of fractions. ) y ) y 5
36 Day Find the derivative of each function. ) y ) y9 ) y ) y y5 9 6) 5) 5 y 7) 8) y 9 y y 0) y 6 9) 5 ) Find the equation of the line tangent to f ( ) at =. Hint: Use product rule & chain rule Find the derivative of each function. (Hint: Rewrite them so they are in terms of negative powers instead of fractions. ) y ) y 6
37 Day Differentiate each function. You may have to re-write, use product rule, quotient rule and/or chain rule. ) y ) g ( ) ) g ( ) ) s ( ) 5) y 6) y *CONTINUE ON NEXT PAGE 7
38 7) y 8) y 5 9) y 0) f ( ) 9 ) y ) f ( ) 9 *CONTINUE ON NEXT PAGE 8
39 ) y ) y 5) gt () t t t 6) f( ) 7) y 8) y *CONTINUE ON NEXT PAGE 9
40 9) 6 5 y 0) y 0
41 Day 6 Find the second derivative of each function. ) f ( ) 5 ) f ( ) 7 ) ) f ( ) f ( ) 5) f ( ) 6) f ( ) 7) f( ) 8) f ( ) 8 *CONTINUE ON NEXT PAGE
42 Find the third derivative of each function 9) f ( ) f( ) 5 0) Find the indicated derivative of each function. ) f '( ) find f ''( ) ) f ''( ) find f '''( ) ) () (6) f find f ( ) ( ) ( ) Find the -value(s) where the second derivative equals zero. f () = 0 ) f ( ) ) f( )
43 Day 7 Find the second derivative of each function. ) f ( ) ) f ( ) ) f ( ) 5 0 ) f( ) f ( ) 6) 5) f ( ) 9 7) f( ) 8) f ( ) *CONTINUE ON NEXT PAGE
44 Find the third derivative of each function 9) f ( ) 0) f( ) Find the indicated derivative of each function. ) f ''( ) 0 6 find f '''( ) ) f find f () '''( ) ( ) ) f find f ( ) ''( ) Find the -value(s) where the second derivative equals zero. f () = 0 ) f ( ) 9 5) f( )
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