192 Calculus and Structures

Transcription

1 9 Calculus and Structures

2 CHAPTER PRODUCT, QUOTIENT, CHAIN RULE, AND TRIG FUNCTIONS Calculus and Structures 9 Copyright

3 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS. NEW FUNCTIONS FROM OLD ONES Given two functions f() and g() we can define three new functions in terms of these old ones: f f ( ) (fg)() = f()g() ; ( )( ) ; and ( f g)( ) f ( g( )) (see Section.6. These new g g( ) functions are called the product, quotient, composition of f and g. If the derivatives of f and g, f () and g (), are known, then we can compute the derivatives of fg, g f, and f () and g (). f g in terms of. THE DERIVATIVE OF THE PRODUCT OF TWO FUNCTIONS To find the derivative of (fg)() use the derivative machine: where, f ( 0 0 h fg)'( ) f ( ) g( ) f ( ) g ( ) mh o( ) () ( 0 0 h ( ) g( 0 ) g'( 0 ) h o( h g ) f ( ) f '( ) h o( ) (a) ) (b) To solve for m, insert Eq. a and b in Eq. to get, ( fg)( ) f ( ) g( ) (( f ( 0 ) ( f '( 0 ) h o ( h))( g( 0 ) g'( 0 ) h o ( h)) f ( 0 ) g( 0 ) (( f ( 0 ) g'( 0 ) f '( 0 ) g( 0 )) h (little o(h) terms) () Eliminate the o(h) terms as usual and, after some algebra, ( fg)'( ) m f ( ) g'( ) f '( ) g( ) (4) Eq. 4 is known as the product rule. Eample: Use the product rule to find the derivative of (fg)() = ( )( ). where f ( ), g ( ) and, f '( ), ( fg )'( ) ( )( ) ( )( ) g' ( )..From Eq. 4, Eample 94 Calculus and Structures

4 Section. Find. d( e ). Let f() = and g() = e where f () = and g () = e. From Eq. 4, d ( e ) e e Problems: a) y= ( 5 )( ) Find: dy/ d(cos ) b) We showed in Sec. 9.6 that sin. Use this with the help of the product rule d ( cos ) to compute:. THE DERIVATIVE OF THE QUOTIENT OF TWO FUNCTIONS The following rule enables you to compute the derivative of the quotient of two functions if you know the derivatives of the two functions.. The proof follows in a similar manner to the Product Rule shown above. I leave the details to the student. ( f f ( ) g( ) f ( )' f ( ) g( )' )'( ) ( )' (5) g g( ) ( g( )) Eq. 5 is known as the quotient rule. Eample Find, d ( ). Let f() = and g() = where f () = and g () =. Replacing these functions in Eq. 5, d( ) ( )() ( ) ( ) - ( ) Calculus and Structures 95

5 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS.4 THE DERIVATIVE OF A FUNCTION OF A FUNCTION THE CHAIN RULE. Consider ( f g)( ) f(g()). I will show by using the Derivative machine that its derivative at 0 is: Using the Derivative machine: ( f g)'( ) f '( g( 0 )) g( 0 ) g( 0 h) g( 0 ) g'( 0 ) h o ( h) f h) f ( ) f '( ) h o ( ) ( h ( f g)( 0 h) f ( g( 0 h)) f ( g( 0 ) g( 0 )' h o ( h )) Set o (h) to zero and let g ( ) h k ' 0 ( f g)( 0 h) f ( g( 0 ) k) = f g( )) f '( g( )) k f ( 0 0 ( g( 0 h)) f ( g( 0 )) f '( g( 0 )) g'( 0 ) h o( h ) Therefore, ( f 0 0 g)'( ) f '( g( )) g( ) (6) Eample 4 Find d Let / f ( ) and g() = +, / f '( ) and g () =.. From Eq. 6, d / ( = ) () Successfully finding the derivative of the function of a function often presents the student with considerable difficulty. There is another approach that I call the bo method which often simplifies this computation. We illustrate this for the case of Eample 4. One first 96 Calculus and Structures

6 Section.4 sketches a diagram of the compound function by representing it as a bo with an outside function and an inside function as shown in Fig.. The function is gotten by placing the inside function, +, into the open parenthesis of the outside function, ( ) / ( ) / + Fig. In order to differentiate the compound function, we use the following step by step procedure: a) Begin with the outside function. Its derivative is: ( ) -/.; b) replace the open parenthesis by whatever is in the bo, i.e., ( + ) -/ ; c) Differentiate the inside d / function, i.e., ; d) multiply the result of b) and c) to get, ( ) (). Eample 5 Sometimes you need several boes in order to define the function. Use the bo method to find, d(cos) First sketch the function. We need three boes to do this. The outermost function is the square function, ( ), the net function is the cosine function, i.e., cos ( ), while the innermost function is the function (see Fig. ). ( ) cos( ). Using a slightly enhanced version of the procedure, Fig. a) d ( ) / = ( ) ; b) Insert everything inside the outer bo into the open parenthesis, i.e., (cos( )) ; c) d cos ( ) / = -sin ( ); d) Insert everything within the second bo into the open parenthesis, i.e., - sin (); e) differentiate the innermost function, i.e., d ()/ = ; f) multiply the results of b, d, and e to get, Calculus and Structures 97

7 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS.5 THE CHAIN RULE d(cos) ()(cos) sin() 9cos ()sin() Computation of the derivative of compound functions is often referred to as the chain rule. Why do we use this terminology? Let us illustrate the chain rule for Eample 4. We can define the compound function by a chain of functions. Eample 6 Define / y ( ) as follows, / y (u) u Therefore we have a chain of dependencies, u y In other words, knowing you can find u and then knowing u you can find y. To find the derivative of y with respect to just differentiate the function down the chain as follows: dy dy du / ( )( ) u () du = where we have inserted the value of u in terms of. Remark : The chain rule merely makes the bo method more formal. Let us apply the chain rule to Eample 5 to find the derivative of. ( cos ) Eample 7 Define y cos () as, y (u) 98 Calculus and Structures

8 Section.5 The chain for this function is: u = cos (w) w = w u y where, dy dy du dw ( )( )( ) u du dw du ( sin w)() cos w( sin w)() 9cos ()sin() where u is epressed in terms of w and then w is epressed in terms of..6 THE INVERSE OF THE CHAIN RULE THE SUBSTITUTION METHOD OF FINDING ANTIDERIVATIVES In Sections.4 and.5 we have computed the derivative of compound functions. In this section we reverse the chain rule and find antiderivatives of certain compound functions using the substitution method. For eample,. Eample 8 Find, We know how to find., ( ) / (7a) /, (7b) and the following procedure that will reduce Epression 7a to 7b. Let s see how it works. a) Let u = + (8) du b) Define, du, or du = c) Solve for, i.e., du d) Replace the result of a) and c) into Eq. 7a, Calculus and Structures 99

9 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS / ( u) du d) Take the constant out of the antiderivate sign using the property b) in Section 0.5, i.e., u / du (9) e) Use the power law (property a) of Section 0.5 to anti differentiate Eq. 9, ( ) u / c (0) f) Epress u in Epression 0 by using Eq. 8, / / ( ) ( ) c. 9 Remark : This substitution method will work whenever the derivative of the substitution, is sitting under d( ) the integral sign differing only by a constant, e.g., in Eample 8,. In other words, the function to be antidifferentiated is off by the constant. Eample 9 Use this procedure to compute, e () a) Let u = b) du du or du = c) du d) u e du or du eu (cancelling ) e) e u c f) e e c 00 Calculus and Structures

10 Section.6 Remark : Since and there is an under the antiderivative sign in epression, so the function to be antidifferentiated differs only by the constant,. Therefore the method will work. Remark 4: We cannot use the substitution method to find the antiderivative of e. (why?).7 TRIG FUNCTIONS The sine and cosine functions of trigonometry are defined in terms of a unit circle. For the unit circle in Fig., if we consider the ray coming from the origin in the direction of the positive - ais as 0 deg. Then (cos,sin ) is the point on the unit circle at a positive angle of deg. measured from the -ais in a counterclockwise direction, and negative angle in the clockwise direction. You can see from Fig., using the Pythagorean Theorem, that sin cos. () cos ( cos θ, sin θ) sin In Section 9. we found that, Fig. d cos sin We can now use this to find the derivative of sin. From the right triangle in Fig. 4 we see that, cos( ) sin, () sin( ) cos. (4) Calculus and Structures 0

11 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS c a Eample 0 d sin Show that, cos. b Fig.4 Using Eq., d sin d cos( ) Using the chain rule, d sin ( )sin( ) cos Eample : Show that d tan sec Let sin tan. cos Using the quotient rule, d tan d sin ( ) cos d tan cos (cos ) ( sin )sin cos Using Eq., d tan sec cos 0 Calculus and Structures

12 Section.6 Eample Show that.7) d tan where tan is the inverse tangent function (see Sec.. and In Section.7 we saw that if y f () then f ( y) f ( f ( )) =, i.e., f ( y) Applying this to y tan we have that tan y In order to find the derivative of an inverse function we state without proof that, dy / dy (5) or, d tan d tan y / dy (6) From Eample, Replacing Eq. 7 in 6, d tan y sec y (7) dy d tan sec y But, y tan so that we have that, d tan (8) (sec(tan )) where sec(tan ) means the secant of the angle whose tangent is. Let s call this angle referring to Fig 5, Calculus and Structures 0

13 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS Fig.5 We see that sec(tan ) And replacing this in Eq. 8 gives, d tan Once we have a new derivative we also have a new antiderivative so that, ( ) tan c 04 Calculus and Structures

14 Chapter Problems Problems Calculus and Structures 05

15 Chapter PRODUCT, QUTIENT, CHAIN RULE AND TRIG FUNTIONS 06 Calculus and Structures

16 Chapter Problems Calculus and Structures 07