1. Package Rule and Product Rule Notes “Tricky Problems”
Derivatives
Package Rule and Product Rule Notes
“Tricky Problems”
By: Benny Teng, 2009~2010

2. Notes Package Rule: y=a n y’=an n-1 ’ Example: f(x)= (5x+1)2
f’(x)= 2(5x+1) = 10(5x+1)

3. Notes Product Rule: y= y’= ’ + ’ Example: f(x)= (2x+1)(5x-3) ’=2 ’=5
’= ’=5
2(5x-3) + (2x+1)(5)
= 10x x+5 = 20x-1

4. f(x)= 7(2x-1)5 , find f’(x) Method 1 package rule f(x)= a n
f’(x)= an n ’
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.

5. f(x)= 7x(2x-1)5 , find f’(x) Product Rule ONLY f(x)= f’(x)= ’ + ’
’= ’ = 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.

6. 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

7. Practice #1
f(x)= 5(1+6x)8, find f’(x)

8. Solution for #1
f(x)= 5(1+6x)8
f’(x) = (8)(5)(1+6x)8-1(6)
= 240(1+6x)7

9. Practice #2
f(x)= 3x(1+x)3, find f’(x)

10. 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)

11. The End