Package Rule and Product Rule Notes “Tricky Problems” Derivatives Package Rule and Product Rule Notes “Tricky Problems” By: Benny Teng, 2009~2010
Notes Package Rule: y=a n y’=an n-1 ’ Example: f(x)= (5x+1)2 f’(x)= 2(5x+1)1 5 = 10(5x+1)
Notes Product Rule: y= y’= ’ + ’ Example: f(x)= (2x+1)(5x-3) ’=2 ’=5 ’=2 ’=5 2(5x-3) + (2x+1)(5) = 10x-6 + 10x+5 = 20x-1
f(x)= 7(2x-1)5 , find f’(x) Method 1 package rule f(x)= a n f’(x)= an n-1 ’ so: 35(2x-1)4 2 = 70(2x-1)4 Method 2 product rule f(x)= f’(x)= ’ + ’ ’ = 0 ’ = 10(2x-1)4 (package rule) so: 0(2x-1)5 7[10(2x-1)4] = 70(2x-1)4 Notice that when 0 cancels out the left side, the remaining is the same as the package rule, so the answer is same as method one.
f(x)= 7x(2x-1)5 , find f’(x) Product Rule ONLY f(x)= f’(x)= ’ + ’ ’= 7 ’ = 10(2x-1)4 (package rule) 7(2x-1)5 + (7x)(10)(2x-1)4 = 7(2x-1)4[(2x-1)+(10x)] = 7(2x-1)4 (12x-1) Why Product Rule Only? Because there is no 0 cancelling out the left side like last example, so if we only use Package Rule, there is still something missing.
Notes when there is no “x”, [ 3(5x-4)4 ] use Package Rule When there is “x” , [ 3x(5x-4)4 ] use Product Rule ONLY No x x
Practice #1 f(x)= 5(1+6x)8, find f’(x)
Solution for #1 f(x)= 5(1+6x)8 f’(x) = (8)(5)(1+6x)8-1(6) = 240(1+6x)7
Practice #2 f(x)= 3x(1+x)3, find f’(x)
Solution for #2 f(x)= 3x(1+x)3 ’= 3 ’=3(1+x)2 f’(x)= (3)(1+x)3 + 3(1+x)2(3x) = 3(1+x)2[(1+x)+(3x)] = 3(1+x)2(4x+1)
The End