DIFFERENTIATION TECHNIQUES CHAIN, PRODUCT & QUOTIENT RULES

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1 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page of 0 MK HOME TUITION Mathematics Revision Guies Level: A-Level Year DIFFERENTIATION TECHNIQUES CHAIN, PRODUCT & QUOTIENT RULES Version : 3 Date:

2 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page of 0 Differentiation Techniques At AS Level, we have ealt with ifferentiation of relatively simple functions Functions like f() = ( 7) ( + ) an g() = full to make them ifferentiable using AS-Level techniques ha to be manipulate or multiplie out in A function like h() = (3 ) 6 woul have ha to have been epane using the binomial series an ifferentiate term by term - a long an rather painful process The Chain Rule or Function of a Function rule Consier the function y = (3 ) 6 How o we ifferentiate it without having to perform a messy binomial epansion? Starting with the variable, we can see that there are two functions; the inner function 3 (to apply to ) an the outer function 6 (to apply to the result of the inner function of 3 ) In function notation, it can be sai that f() = 3 ; g() = 6 ; gf() = (3 ) 6 Alternatively, y = (3 ) 6 can be epresse as a two-stage function; the intermeiate function u = 3 followe by y = u 6 The erivative of the combine function can be evaluate by using the chain rule, or, in function notation, g( f()) g( f()) f () More links are possible in the chain, such as in usually come up in eams! Eample (): Use the chain rule to ifferentiate y = (3 ) 6 Let u = 3 y = u 6 6 6u 6(3 ) (Substitute 3 for u) 6(3 ) (6) 36 (3 ), but such eamples will not

3 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page 3 of 0 Eample (): Use the chain rule to ifferentiate y = (8 ) Here, u = 8 y = u u (8 ) (8 ( ) ) (8 ) With practice, the intermeiate working can be one mentally Eample (3): Use the chain rule to ifferentiate y = (7 + ), without showing the sie working If you ifferentiate ( thing ), you have ( thing ) erivative of thing (7 ) 7 3(7 ) Notice how the multiplier of 3 came about; it is the original power of the epression () multiplie by the coefficient of the inner -term in the brackets (7) This hols true for all epressions of the type (a + b) n where the inner function is linear in : ( a b) n an( a b) n (This also inclues fractional an negative n) Eamples (): Without using sie working, use the chain rule to fin : i) i) ( 3 ) ; ii) ( 3) ( 3) ( 3 ) ; iii) 8 ii) ( 3) = ( ( 3) 0( 3) 3 3), an therefore or 0 ( 3) 3 iii) 8 (8 ), an so (8 ) (8 ) or 8

4 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page of 0 The Proct Rule At AS level, we ifferentiate y = ( 7)( + ) by epaning it as + 8, an then ifferentiating it to obtain y= + (yis another way of writing ) A proct of two functions can be ifferentiate by using the following rule: If y = uv, where u is a function f() an v is another function g(), then u v or in function notation, ( f() g() )= f() g() + f () g() Eample (): Differentiate y = ( 7)( + ) using the proct rule Let u = 7 an v = + an u v ( 7)( + )() = = + The erivative of the proct is (first erivative of secon) + (secon erivative of first) Eample (6): Differentiate y = ( - 3) ( + ) using the proct rule an simplify the result Let u = - 3 an v = + an = ( - 3) () + ( + ) ( - ) (first erivative of secon) + (secon erivative of first) = = If the question oes not ask for the result to be simplifie, then the last two steps can be omitte

5 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page of 0 Eample (7): Differentiate y = 7 3 using the proct rule, simplifying the result Hence show that the graph of y has no turning points Although this result looks like a quotient rather than a proct, we can reefine the epression as the proct of u = - 7 an v = 3 From this, we work out 0 7 an Therefore: 3 3 by the chain rule (working not shown) u v (0 7)(3 ) = = = 3 3 For the graph of y to have turning points, the graient must equal zero for some value(s) of The quaratic numerator of the erive function, however, has a iscriminant of (-0) ( 8), or -80, which is negative, implying no real roots, ie no zero graient anywhere

6 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page 6 of 0 The Quotient Rule At AS level, we ifferentiate y = y 3 by rewriting it as, to obtain the result If y = v u, where u is a function f() an v is another function g(), then v u v, or or in function notation, f ( ) g( ) g( ) f ( ) f ( ) g( ) ( g( )) Eample (7): Differentiate y = using the quotient rule, an simplify the result Let u = + an v = 3 an v u 3 ( ( )) (( v )()) ( ) = = 3 Again, the last steps can be omitte if the question oes not ask for the result to be simplifie

7 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page 7 of 0 Eample (8): Differentiate y = 3 using the quotient rule, an simplify the result Let u = 3 + an v = - 6 an The orer of the work is: square the bottom line of the epression to give the bottom line of the result, then write the bottom erivative of top top erivative of bottom on the top line of the result ( )(6) (3 ( ) )() ie Simplifying the top line gives 3 ( ) Eample (9): Differentiate y = (This is a repeat of Eample (6)) 7 3, using the quotient rule an simplify the result Let u = 7 an v = an 3 v u v (3 )(0 7) ( (3 ) 7)(3) (30 6 8) ( 3 ) 0 8 3

8 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page 8 of 0 Using the result = / This is another result that is useful for ifferentiating inverse functions, especially inverse trigonometric functions to be iscusse in later stu Eample (0): Differentiate y = by using an fin its graient at the point (, ) How is the result relate to the graient of y = at the point (, )? First, rewrite the epression as = y Differentiation gives y an therefore y Finally, we must rewrite the erivative in terms of, so substituting for y gives This is equivalent to The graient of y = at (, ) is or 0 The proct of the graients is y 0, whilst the graient of y = at the point (, ) is (Do not confuse this with the rule for perpenicular lines, whose graients have a proct of ) If the graph of f() passes through the point (p, q) an its graient at that point is equal to m, then the graient of the graph of the inverse function f -- () is equal to m at the point (q, p)

9 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page 9 of 0 More on the chain rule relate rates of change The chain rule can also be use to establish results for relate rates of change Eample (): Crue oil leaking out of a tanker proces a circular slick with raius r km an area A km The raius of the slick increases with time at a rate given by r = 0 km/h t Fin the relate rate of change of area of the slick, A, an also the rate of increase of the area of the t slick at a time when the raius is km The area of a circle is given by A = r The change in area with respect to the change in raius is therefore A = r r By the chain rule, A t A r r t Here it is r 0 or r At the time when the raius of the slick is km, the rate of increase in the slick s area is km /h or about 6 km /h

10 Mathematics Revision Guies Differentiation: Chain, Proct an Quotient Rules Page 0 of 0 Eample (): During a tunnel-builing project, it is foun that the volume V of the ecavate earth is relate to the height h of the resulting pile by the formula 6 V = h h i) Fin the value of V at the instant when the pile of earth is m high h ii) The volume of the pile of ecavate earth is increasing at a constant rate of 0 m 3 per hour Fin the rate (per hour) at which the height of the pile is increasing at the instant the pile is m high Give the final result to the nearest centimetre (Copyright OCR, GCE Mathematics Paper 73, January 008, Q, altere) i) By the chain rule, u = h 6 + h an V = u Therefore 6h an h V u (or V u ) Hence V h V h V h 6h ( h 6 h ) Substituting h = into the above erivative gives V h 6(3) (6 ) V h 93 6 ii) From the given ata, we ece that the change (per hour) in volume of the earth, or 0 h We want to fin the rate of change of height when the pile is m high, or, when h = t V t h Now = t h V V t by the chain rule, so at h =, from (i), t h 6 = 0 or Note that h V V h At the instant when the pile of ecavate earth is m high, the height is increasing at a rate of 83 cm per hour