Derivatives of products and quotients

Derivatives of Products and Quotients!!

Product Rule!

If f(x) = r(x)^. s(x) and if both derivatives exist, then
f '(x) = r(x) ^. s'(x) + s(x) ^. r'(x)
In words, this means the derivative of a product is the first function times the derivative of the second function plus the second function times the derivative of the first function!!

Sample problems.

1) f(x) = 3x²(5x² - 3x) Find f'(x) using the product rule.

Solution:

r(x) = 3x²	s(x) = 5x² - 3x
r'(x) = 6x	s'(x) = 10x - 3

Set up a chart like the one above. Then cross multiplying the values in the chart will give you f'(x)

f'(x) = 3x²(10x - 3) + 6x(5x² - 3x)
= 30x³ - 9x² + 30x³ - 18x²
= 60x³ - 27x²

2) f(x) = (2x - 1)(3x + 2). Find f '(x) by using the product rule!

Solution:

r(x) = 2x - 1	s(x) = 3x + 2
r'(x) = 2	s'(x) = 3

Use the chart and cross multiply again to find f '(x)!!
f '(x) = 3(2x - 1) + 2(3x + 2)
= 6x - 3 + 6x + 4
= 12x + 1

Solution:

r(x) = x^1/2 + 2	s(x) = x² - 3x
r'(x) = 1/2x^1/2	s'(x) = 2x - 3

Quotient Rule!

If f(x) = r(x)/s(x) and all derivatives exist, and if s(x) does not equal 0, then

f '(x) = [s(x) ^. r'(x) - r(x) ^. s'(x)]/[s(x)]²

In words, this means the derivative of a quotient is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all over the square of the denominator.
(or lo-di-hi, minus hi-di-lo, over lo-lo!!) Figure this one out!!!

Sample problems

Solution:

r(x) = 2x - 1	s(x) = 3x + 4
r'(x) = 2	s'(x) = 3

Use the chart like we did for the product rule, but it is important you put them in the correct order. Use the cross product from the left bottom corner and upper right corner first!

Find f '(x).
Solution:

The problem has both the product and quotient rule. We need to apply both. First find the derivative of the numerator by using the product rule!

r(x) = 3 - 4x	s(x) = 5x + 1
r'(x) = -4	s'(x) = 5

The derivative of the numerator is:
5(3 - 4x) + (-4)(5x + 1) =
15 - 20x - 20x -4 =
11 - 40x
Now use the quotient rule!

r(x) = (3 - 4x)(5x + 1)	s(x) = (7x - 9)
r'(x) = 11 - 40x	s'(x) = 7

Use the above chart to find f '(x)
f '(x) = [(-40x +11)(7x - 9) - 7(-4x +3)(5x + 1)]/(7x - 9)²
= [(-280x² + 360x + 77x - 99) - 7(-20x² - 4x + 15x + 3)]/(7x - 9)²
= [(-280x² + 437x - 99) + (140x² + 28x - 105x - 21)]/(7x - 9)²
= (-140x² +360x - 120)/(7x - 9)²

On to the chain rule!!