Section 3 4A: The Chain Rule Introuction The Power Rule is state as an x raise to a real number If y = x n where n is a real number then y = n x n-1 What if we wante to fin the erivative of a variable expression raise to a real number where the variable expression is NOT x? Fin y if y = ( 3x +1) 2 ( 3x +1) 2 oes not fit the requirement of the power rule that the base be x To use the Power Rule we coul FOIL out ( 3x +1) 2 to get 3 separate terms each which each fit the power rule. Fin y if y = ( 3x +1) 2 y = 9x 2 + 6x +1 y = 18x + 6 y = 6 ( 3x +1) If we try to use the power rule on a problem where the variable expression is NOT x will it prouce the correct answer? Fin y if y = ( 3x +1) 2 y = 2 ( 3x +1) ( ) we got the answer of y = 6 3x +1 by FOILing an then useing the Power Rule for a base of x. It is off by a factor of 3 The variable expression was 3x + 1 not x the erivative of 3x + 1 is 3 Using The Power Rule on a problem where the base is not x will NOT prouce the correct answer? Math 400 3 4A Chain Rule Page 1 of 5 2013 Eitel
The erivative rule for y = e x is state for e raise to the x power If y = e x then y = e x e 2x Fin y if y = e 2x oes not fit the requirement that the exponent of base e be x To use the erivative rule for y = e x we must express e 2x as e x e x to get a prouct of 2 terms an then use the prouct rule Fin y if y = e 2x y = e x e x the 67 first 8 er 64 of 7 the 48 sec the 64 7 secon 48 er 64 of 7 the 48 first y = e x e x + e x e x y = e 2x + e 2x y = 2e 2x If we try to use the erivative rule for y = e x on a problem where the exponent is not x will it prouce the correct answer? Fin y if y = e 2x y = e 2x the answer of y = 2e 2x we got useing the Prouct Rule It is off by a factor of 2 Using the erivative rule for y = e x The exponent above e was 2x not x the erivative of 2x is 2 on a problem where the base is not x will NOT prouce the correct answer? Math 400 3 4A Chain Rule Page 2 of 5 2013 Eitel
The erivative rule for y = sin x is state for the sin of x If y = sin x then y = cos x Fin y if y = sin(2x) 2x oes not fit the requirement that the variable expression be x To use the erivative rule for y = sin 2x we must express y = sin 2x as y = 2 sin cos x to get a prouct of 2 terms an then use the erivative rule for y = sin x Fin y if y = sin(2x) y = 2 sin x cos x 6 the 47 first 48 6 er 4 of 7 the 4 sec 8 6 the 4 secon 748 y = 2sinx sin x + cos x y = 2sin 2 x + 2cos x 2 6 er 4 of 7 the 4 first 8 2cosx y = 2 ( sin 2 x + cos 2 x) Note: sin 2 x + cos 2 x = cos(2x) y = 2cos(2x) If we try to use the erivative rule for y = sin x where the variable expression is NOT x will it prouce the correct answer? Fin y if y = sin(2x) y = cos(2x) the answer of y = 2cos2x we got using the Prouct Rule It is off by a factor of 2 The variable expression is 2x an the erivative of y = sin (2x) was off by a factor of 2 Using The rule for the erivative y = sin x on a problem like y = sin (2x) will NOT prouce the correct answer? Math 400 3 4A Chain Rule Page 3 of 5 2013 Eitel
It is clear from the examples above that you CANNOT use the erivative rules that require the variable to be exactly x on functions with expressions whose variable is NOT exactly x. Are we require to transform each function into variable expression ion terms of only x? NO: There is a way to use the simple rules for the erivatives of the basic functions but to o so we nee to unerstan how to ajust the final answer to take in to account what aing expressions whose variable is NOT exactly x has on those rules. The Chain Rule The function f (x) = x 2 is relate to the function y = (3x +1) 2 The erivative of f (x) It is written as f (x). It is an expression of the change in f (x) with respect to x. It is written as ( x f (x)). The function f (x) = (3x +1) 2 is the composite function f ( g(x) ) where f (x) = x 2 an g(x) = (3x +1) To avoi confusion between the ifferent functions we rewrite the composite function ( ) = (3x +1) 2 where f (x) = x 2 an g(x) = (3x +1) f g(x) to rea f (u) = u 2 where u = (3x +1) The erivative of f ( u) is an expression of the change in f (x) with respect to u times the change in u with respect to x. It is written as f (x) = ( u f (u)) x u To fin x (3x +1)2 must be thought of a fin u [ ] u2 times u x where u = ( 3x +1 ) f (x) = ( ) u u2 (3x +1) x ( ) = 2u n 1 u u2 (3x +1) = 3 x n } u n 1 (3x +1) 64 748 6 x 47 48 f (x) = 2 (3x +1) 1 3 f (x) = 6(3x +1) It is the aition of the prouct of the erivative of u = (3x +1) what makes the erivative of y = (3x +1) 2 ifferent from the erivative of simply y = (x) 2 Math 400 3 4A Chain Rule Page 4 of 5 2013 Eitel
The Constant Multiple Rule x [ c u] = c u The Prouct Rule x [ u v] = u v v u The Sum an Difference Rule x [ u ± v ±...] = u ± v... The Quotient Rule x u v = v u u v v 2 The Power Rule in terms of x The Power Rule where x is some f (x) = u 1. x [ ] xn = n x n 1 1. 2. x [ ] ex =e x 2. 3. x [ ln(x) ] = 1 x 3. x (u)n = u n 1 u x [ ] eu =e u u x [ ln(u) ] = 1 u u = u u 4. [ x log b (x)] = 1 ln(b) x 4. x log b (u) [ ] = u ln(b) u 5. x [ ] ax = ln(a) a x 5. x [ ] au = ln(a) a u u Trig Rules 6. 7. x [ sin x] = cos x 6. x [ sin u] = cos u u x [ cos x] = sin x 7. x [ cos u] = sinu u 8. x tan x [ ] = sec2 x 8. ( ) u x tan u [ ] = sec2 u ( ) 9. 10. x [ sec x] = sec x tan x 9. x [ csc x] = csc x cot x 10. [ ] = ( sec x tanx ) ( u ) x sec u [ ] = ( csc u cot u ) ( u ) x csc u 11. x cot x [ ] = csc2 x 11. ( ) u x cot u [ ] = csc2 u ( ) Math 400 3 4A Chain Rule Page 5 of 5 2013 Eitel