Section 3 4B Lecture The Chain Rule If y = f (u) is a differentiable function of u and u = g(x) is a differentiable function of x then dy dx = dy du du dx or If y = f (u) then y = f (u) u The Chain Rule with the Power Rule Theorem: If a and n are a real numbers If y = a u n then y = a n u n-1 u or in English If y = a ( some function of x inside the brackets ) n then y = a n ( some function of x inside the brackets ) n-1 the derivative of the function inside the brackets Example 1 The Chain Rule with the Power Rule Find y if y = ( 5x 2) 3 if we let u = 5x 2 then y = ( u) 3 where u = 5x 2 and u = 5 y = n(u) n 1 u n y = 3 u n 1 " $ # $ % ( 5x 2) 2 u 5 y = 15( 5x 2) 2 Math 400 3 4B Lecture Chain Rule Page 1 of 11 2019 Eitel
Example 2 The Chain Rule with the Power Rule Find y if y = 2( 3x 2 + 5x) 4 Think of the outside function as 2(u) 4 take the derivative of the outside function 2 4(u) 3 3 8 3x 2 + 5x now multiply that by the derivative of the inside function if u = 3x 2 + 5x then u = 6x + 5 a n " # 2 4 u n 1 $ " $ # ( 3x 2 + 5x) 3 $ " u $ # ( 6x + 5) 8( 3x 2 + 5x) 3 ( 6x + 5) Example 3 The Chain Rule with the Power Rule Find y if y = 4( tan x) 3 Think of the outside function as 4(u) 3 take the derivative of the outside function 4 3(u) 2 12( tan x) 2 now multiply that by the derivative of the inside function if u = tan x then u = sec 2 x a n " $ # $ % " $ # u $ % 12 tan x sec 2 x u n 1 2 12tan 2 x sec 2 x Example 4 Math 400 3 4B Lecture Chain Rule Page 2 of 11 2019 Eitel
The Chain Rule with the Power Rule Find y if y = 3 ( 6x 2 1) = ( 6x 2 1) 1/3 Think of the function as (u) 1/3 take the derivative of the outside function 1 3 (u) 2/3 2/3 1 3 6x2 1 now multiply that by the derivative of the inside function if u = 6x 2 1 then u = 12x n " $ u# n 1 $$ % 1 ( 6x 2 1) 2/3 12x 3 "#% u 1 3 12x ( 6x2 1) 2/3 4x ( 6x 2 1) 2/3 Example 5 The Chain Rule with the y = sin (u) Find f if f (x) = sin( 3x 4 ) if y = f (u) then y = f (u) u Think of the outside function as sin(u) take the derivative of the outside function to get cos(u) cos 3x 4 now multiply that by the derivative of the inside function if u = 3x 4 then u = 12x 3 f (u) # "# $ cos 3x 4 % u 12x 3 12x 3 cos( 3x 4 ) Math 400 3 4B Lecture Chain Rule Page 3 of 11 2019 Eitel
Example 6 The Chain Rule with y = e u Find f (x) if f (x) = e x 2 if y = f (u) than y = f (u) u Think of the outside function as e u take the derivative of the outside function to get e u e x 2 now multiply that by the derivative of u if u = x 2 then u = 2x f (u) e x u 2 2x 2x e x 2 Example 7 Find f (x) if f (x) = e x = e x1/2 if y = f (u) than y = f (u) u Think of the outside function as e u take the derivative of the outside function to get e u e x1/2 now multiply that by the derivative of u if u = x 1/2 then u = 1 2 x 1/2 f (u) e x1/2 "# u $ 1 2 x 1/2 e x 2x 1/2 e x 2 x Math 400 3 4B Lecture Chain Rule Page 4 of 11 2019 Eitel
Example 8 The Chain Rule with y = ln( u) Find y if y = ln( 3x 4 + 2x 3 ) if u = 3x 4 + 2x 3 then u = 12x 3 + 6x 2 if y = ln(u) then y = u u y = 12x3 + 6x 2 3x 4 + 2x 3 y = 6x2 2x +1 x 3 3x + 2 y = 6 2x +1 x 3x + 2 Example 9 The Chain Rule with y = ln( u) Find y if y = ln y = ln 3x 2 5x 3x 2 5x 1/2 = 1 2 ln ( 3x2 5x) if u = 3x 2 5x then u = 6x 5 if y = ln(u) then y = u u y = 1 2 6x 5 3x 2 5x y = y = 6x 5 2 3x 2 5x 6x 5 2x 3x 5 Math 400 3 4B Lecture Chain Rule Page 5 of 11 2019 Eitel
Example 10 The Chain Rule with y = ln u = ln u lnv v x 2 1 Find y if y = ln sin(2x) Use the law of logs: log u v y = ln ( x 2 1) ln ( sin(2x) ) if y = ln(u) then y = u u = log( u) log( v) der of sin(2x) der ofx 2 y = 2x " $ # $ % x 2 1 cos(2x) 2 sin2x y = 2x x 2 1 2 cos(2x) sin(2x) The Chain Rule with y = ln Example 11 u i w v = ln u i w Find y if y = ln x sin(2x) ln( v) = ln( u) + ln( w) ln( v) e x2 y = ln ( x sin(2x) ) ln e x2 y = ln(x) + ln(sin(2x)) x 2 ln e y = ln(x) + ln(sin(2x)) x 2 if y=ln(u) then y = u u der of x der of sin(2x) y = 1 " $ # $ % x + cos(2x) 2 sin(2x) 2x der of x 2 y = 1 x + 2 cos(2x) sin(2x) 2x Math 400 3 4B Lecture Chain Rule Page 6 of 11 2019 Eitel
Repeated use of the Chain Rule then y = f g If y = f ( g[ ( h{ x} ) ] ) ( [ ( h{ x} ) ] ) g [ ( h{ x} ) ] h x ( { } ) Repeated use of the Chain Rule Example 12 ([ ]) Find f (x) if f (x) = sin cos 4x d dx ([ ]) sin cos 4x ([ ]) d dx cos cos 4x cos ( 4x ) cos cos 4x cos cos 4x ([ ]) sin(4x) d dx 4x ( ) ( sin(4x) )( 4 ) 4 cos cos 4x ( ) sin ( 4x ) an alternate notation for example 12 ( ) Find y if y = sin cos 4x if u = cos 4x y = sin u then where u = cos 4x the der #" of # sin # $ u the der #" of # cos $ 4x y = cos cos 4x sin 4x ( ) y = 4 cos cos 4x ( ) sin ( 4x ) the der % of 4 x 4 Math 400 3 4B Lecture Chain Rule Page 7 of 11 2019 Eitel
Example 13 Repeated use of the Chain Rule Find f (x) if f (x) = ( cos ( 4 x )) 3 d dx ( cos ( 4x ))3 ( ) 2 d dx 3 cos 4x cos ( 4x ) 3 cos 4x ( ) 2 sin(4x) d dx 4x 3 cos 4x ( ) 2 sin(4x) 12 cos 4x 4 ( ) 2 sin 4x an alternate notation for example 13 Find y if y = ( cos ( 4x )) 3 if u = cos ( 4x ) then y = u 3 where u = cos 4x the # der " of # (u) # $ n ( ) 2 y = 3 cos 4x y = 12 cos 4x the der # of " cos(4 # $ x) sin 4x ( ) 2 sin 4x the der % of 4x 4 Math 400 3 4B Lecture Chain Rule Page 8 of 11 2019 Eitel
Example 14 The Chain Rule with the Product Rule Find f (x) if f (x) = 4x 3x 2 2 = 4x ( 3x 2 2) 1/2 " $ $ der # of $ sec$ $ % the first 1 4x 2 3x2 2 1/2 6x " $ the # sec $ % 1/2 + 3x 2 2 der of first 4 12x 2 ( 3x 2 2) 1/2 + 4( 3x 2 2) 1/2 factor out a ( 3x 2 2) 1/2 1/2 12x 2 + 4( 3x 2 2) 3x 2 2 [ ] [ ] 1/2 12x2 +12x 2 8 3x 2 2 24x2 8 3x 2 2 Math 400 3 4B Lecture Chain Rule Page 9 of 11 2019 Eitel
Example 15 The Chain Rule with the Quotient Rule Find f (x) if f (x) = 2x 3x 2 2 2x = ( 3x 2 2) 1/2 the # bottom "# $ ( 3x 2 2) 1/2 der % of top the % top # der # of " bottom ### $ 2 2x 1 ( 2 3x2 2) 1/2 6x ( 3x 2 2) 2 &# '#( bottom squared 2 3x2 2 1/2 6x 2 ( 3x 2 2) 1/2 ( 3x 2 2) 2 factor out a ( 3x 2 2) 1/2 [ 6x 2 ] ( 3x 2 2) 2 ( 3x 2 2) 1/2 2 3x 2 2 [ ] 5/2 6x2 4 6x 2 3x 2 2 4 ( 3x 2 2) 5/2 Math 400 3 4B Lecture Chain Rule Page 10 of 11 2019 Eitel
Example 15 The Chain Rule with the Power Rule and the Quotient Rule Find f (x) if f (x) = x2 2 2 2x + 3 the # der " of # u$ n the # bottom "# $ 2 x2 1 2 ( 2x + 3) 2x + 3 % # the " top # $ x 2 2 der of top 2x ( 2x + 3) 2 &# '#( bottom squared der of % bottom 2 2 x2 2 2x + 3 4x2 + 6x 2x 2 + 4 2x + 3 2 2 x2 2 2x + 3 2x2 + 6x + 4 2x + 3 2 ( 4x 2 +12x + 8) x2 2 2x + 3 3 Math 400 3 4B Lecture Chain Rule Page 11 of 11 2019 Eitel