The Chain Rule. This is a generalization of the (general) power rule which we have already met in the form: then f (x) = r [g(x)] r 1 g (x).

Transcription

1 The Chain Rule This is a generalization of the general) power rule which we have already met in the form: If f) = g)] r then f ) = r g)] r g ). Here, g) is any differentiable function and r is any real number. It may be easier to remember it in the form The derivative of ) r is r ) r Derivative of what is in ) You may put any differentiable epression inside ). Of course if you put in, you get what you already know, that the derivative of r is r r. To get something new, you have to put in epressions like +, 2, 4 2, etc. This etends to the trigonometric and the eponential functions as follows:. The derivative of sin ) is cos )] Derivative of what is in ). In practice, it is more convenient to write it as The derivative of sin ) is Derivative of what is in )] cos )]. If you put inside ), you get what you already know, that the derivative of sin is cos.) Eample a) Given f) = sin 6, view it as f) = sin6). Then f ) = 6] cos 6] = 6 cos 6. b) Let g) = sin 2 ). Then g ) = 2 cos 2 ). 2. The derivative of cos ) is Derivative of what is in )] sin )]. Eample 2 a) Given f) = cos 2, view it as f) = cos 2 ). Then f ) = 2)] sin 2 )] = 2 sin 2. b) Let g) = cos + 2 ). Then g ) = ) sin + 2 ). The derivative of tan ) is Derivative of what is in )] sec 2 ) ]. Eample a) Given f) = tan 4, view it as f) = tan4). Then f ) = 4] sec 2 4) ] = 4 sec 2 4. b) Let g) = tan ). Then g ) = + ) sec 2 2 ). 4. The derivative of csc ) is Derivative of what is in )] csc ) cot )].

2 Eample 4 a) Given f) = csc 7, view it as f) = csc7). Then f ) = 7] csc 7) cot 7)] = 7 csc 7 cot 7. b) Let g) = csc 2 ). Then g ) = 2 csc 2 ) cot 2 ) 5. The derivative of sec ) is Derivative of what is in )] sec ) tan )]. Eample 5 a) Given f) = sec 9, view it as f) = sec9). Then f ) = 9] sec 9) tan 9)] = 9 sec 9 tan b) Let g) = sec + 2 ). Then g ) = 6 ) sec + 2 ) tan + 2 ) The derivative of cot ) is Derivative of what is in )] csc 2 ) ]. Eample 6 a) If f) = cot 2 ) then f ) = 2] csc 2 2 )] = 2 csc2 2. b) If g) = cot 5 ) then g ) = ) csc 2 5 ). 7. The derivative of e ) is Derivative of what is in )] e )]. Eample 7 a) Given f) = e 4, view it as f) = e 4). Then f ) = 4] e 4)] = 4e 4. b) If g) = e sin then g ) = cos ] e sin ] = e sin cos The following is a summary of what we have described: Function Derivative. ) n n ) n Derivative of what is in ) 2. sin ) Derivative of what is in )] cos )]. cos ) Derivative of what is in )] sin )] 4. e ) Derivative of what is in )] e )] 5. tan ) Derivative of what is in )] sec 2 ) ] 6. cot ) Derivative of what is in )] csc 2 ) ] ) 7. sec ) Derivative of what is in )] sec ) tan )] 8. csc ) Derivative of what is in )] cot ) csc )] 2

3 Quite often, you may have to use the chain rule more than once. For an eample, take f) = sin. If we write sin as sin ) /2 then the derivative of f is 2 sin ) /2 Derivative of sin ) We have to use the chain rule a second time to find the derivative of sin. This turns out to be cos ). Therefore f ) = 2 sin ) /2 cos ) = 2 sin ) /2 cos ) Eercise 8. Determine the derivative of each function and simplify it when possible. a) f) = 2 sin c) v) = 2 sec 4 7 b) w) = 2 sin cos 2 d) f) = csc 2 e) g) = e 2 f) h) = 2 tan g) u) = 4 cos sec ) h) u) = 2e cot 5 i) f) = 4 7 tan j) g) = 2 tan k) h) = 4 2 sin 2 l) u) = sec 2 m) v) = sin n) w) = cos + 2 o) f) = e 2 sin cos ) p) g) = cot q) h) = tan 2 csc ) 2 2 r) u) = e +sec 2 2 The Chain Rule as a Rule for Computing Derivatives of Compositions The chain rule, in its general form, is a rule for computing derivatives of compositions of functions. It provides us with a method of computing the derivative of a composition f g of differentiable functions f and g if we know the derivative of g at a point and the derivative of f at g). More precisely, it states that the derivative of f g at is given by the formula f g) ) = f g)) g ). An outline of a proof is at the end of this section.) The results in table on page 2 are all special cases of this general form. Here is an illustration: Eample 9 To find the derivative of h) = sec 2 + ) using the formula f g) ) = f g)) g ), write h) as the composition h) = f g) where g) = 2 + and f) = sec. Then f g)) = sec 2 + ) tan 2 + ) and g ) = 2 +. Therefore h ) = f g)) g ) = 2 + ) sec 2 + ) tan 2 + )] For comparison to the result of using formula 7) in table, write h) as sec ). The term sec ) tan )] is sec 2 + ) tan 2 + )], which corresponds to f g)). The term Derivative of what is in ) is 2 + ), which corresponds to g ). Table cannot be used to find the required derivative in the net eample: Eample 0 To find f g) 2) given that g2) =, g 2) = 8 and f ) = 4. By the chain rule, f g) 2) = f g2))g 2) = f )g 2) = 4 8 = 6

4 Eercise. In the figure below, you are given the graph of a function f and the tangents to its graph at the two points with -coordinates 2 and.5. Let g) = Use the chain rule and the given tangents to estimate f g) 2) and f g) ) Use the chain rule, in the form f g) ) = f g)) g ), to find the derivative of the given function. In each case, you should write the given function as a composition of functions f and g of your choice. a) h) = b) u) = 2 + ) 4 c) v) = 2 cos +5 cos2 Another way of applying the chain rule We illustrate with eamples: Eample 2 Let h) = tan ). We reduce it to a more familiar function by introducing a new variable u =. Then h) = tan u. Let fu) = tan u. We can now write h) = fu) = tan u. The derivative of f, its variable is u), is df du = sec2 u. The derivative of u, its variable is ), is du d = 2. The chain rule states that the derivative of h is dh d = df du du d = sec 2 u ) 2) = 2 sec 2 ) As you have probably figured out already, this is similar to writing h) as a composition h) = f g) where g) = and f) = tan. The only difference here is that we have written g) as a new variable u; therefore we must write fg)) as fu). In the epression df du du for the derivative of h, d df du is really f g)) and du d is g ). Eample Let h) = 4 csc 2) + sec 2 4 csc u + sec u. Clearly, derivative of h is ). To find its derivative, we let u =. Then h) = fu) = 2 d =. By the chain rule, the 22 df du = 4 csc u cot u + sec u tan u and du dh d = df du du d = 4 csc u cot u + sec u tan u) ) 2 2 = 4 csc 2 2) cot 2) sec 2) tan )] 2 2 4

5 Remark 4 The chain rule, in the form of table, is easier to use in reverse to guess antiderivatives of functions. Eercise 5 Find the derivative of each function: i) h) = 7e 2 ii)v) = 4 tan iii) w) = e tan iv) z) = 2 + ) Sketch of a Proof of the Chain Rule Let f and g be differentiable functions, f g be their composition and c be a point in the domain of g. The derivative of g at c is the number g c) given by g c) = lim c g) gc) c The derivative of f at gc) is the number f gc)) given by f gc)) = fu) fgc)) lim u gc) u gc) The derivative of f g) at c is the number f g) c) given by fg)) fgc)) Now write as c tools, it can be shown that f g) c) = lim c fg)) fgc)) c fg)) fgc)) g) gc) and denote g) by u. Then, with appropriate g) gc) c f g) fg)) fgc)) g) gc) c) = lim = c g) gc) c lim c ) fu) fgc)) u gc) fu) fgc)) fu) fgc)) It can also be shown that lim = lim. Therefore c u gc) u gc) u gc) ) ) f g) fu) fgc)) g) gc) c) = lim lim = f gc))g c) u gc) u gc) c c lim c ) g) gc) c In general, if g is differentiable at and f is differentiable at g), then f g is differentiable at and You are asked to give another proof in the eercises. Eercise 6 f g) ) = f g))g ) 2). Determine the derivative of each function and simplify when possible. a) f) = cos sin ) 2/ b) g) = tan ) c) h) = cos 2 d) u) = csc sec e) v) = sin 2 f) w) = 4 sec 2 g) f) = 4 cot j) u) = sin 5 ) ) Let f) =. Show that f ) = 2 +. Let g) =. Show that g ) = h) g) = e 2 sin 2 i) h) = + cot 4 k) v) = csc ) l) w) = tan )

6 a sin + b cos 4. Let a, b, c, and d be constants. Show that f) = c sin + d cos has derivative f ad bc ) = c sin + d cos ) Copy and fill in the missing details in the following sketch for the proof of the chain rule: Let f and g be differentiable functions, f g be their composition and c be a point in the domain of g. The derivative of g at c is the number g c) with the property that g) = gc) + c) g c) + c) b ) ) where b ) is a function whose values approach 0 as approaches c. Likewise, the derivative of f at gc) is the number f gc)) with the property that fu) = fgc)) + u gc)) f gc)) + u gc)) b 2 u) 4) where b 2 u) is a function whose values approach zero as u approaches gc). By substituting ) into 4), show that fg)) = fgc))+ c) g c) + c) b )] f gc))+ c) g c) + c) b )] b 2 g)) This simplifies into fg)) = fgc)) + c) f gc))g c) + c) ] b )f gc)) + g c)b 2 g)) + b )b 2 g)) But g) approaches gc) as approaches c, therefore b 2 g)) approaches 0 as approaches c. Show hat the epression b )f gc)) + g c)b 2 g)) + b )b 2 g)) inside the above square brackets approaches 0 as approaches c. Denote it by b ) and write fg)) as f g) to conclude that f g) = fgc)) + c) f gc))g c) + c) b ) 5) where b ) approaches 0 as approaches c. But we know that the derivative of f g at c is the number denoted by f g) c) with the property that f g) = fgc)) + c) f g) c) + c) b) 6) where b) approaches 0 as approaches c. Compare 5) and 6), and deduce that the second term in 5) must be identical to the second term in 6). Therefore f g) c) = f gc))g c) 6