`(d/dx)( f(x)g(x) ) = f(x)( (d/dx)g(x) ) + g(x)( (d/dx)f(x) )`

What this means is that the derivative of a product equals the first factor times the derivative of the second factor, plus the second times the derivative of the first.

Proof:
Letting `y = f(x)g(x)` and using the definition of a derivative:
`f(x+h) - f(x)
`lim -------------
`h→0 h`
we write
`dy f(x+h)g(x+h) - f(x)g(x)
-- = `lim` ------------------------.
dx h→0 h`
To change the form of the quotient so that the limit may be evaluated, we subtract and
add the expression `f(x+h)g(x)` in the numerator.
`
dy f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) - f(x)g(x)
-- = lim ------------------------------------------------- =
dx h→0 h
_ _
| g(x+h) - g(x) f(x+h) - f(x) |
`lim|`f(x+h) * -------------- + g(x) * -------------- |=
|h→0 h h |
|_ _|
g(x+h) - g(x) f(x+h) - f(x)
`lim` f(x+h) * `lim` ------------- + lim g(x) * lim ------------
h→0 h h
`

Since

`f` is differentiable at

`x`, it is continuous at

`x`. Hence, lim

_{h→0} f(x+h) = f(x). Also, lim

_{h→0} g(x) = g(x),
since

`x` is fixed in this limiting process. Finally, applying the definition of derivative to

`f(x)` and

`g(x)`, we obtain

dy

-- = f(x)g'(x) + g(x)f'(x).

dx
The product rule is good to use if the factors of the product are too complicated to multiply out beforehand. However, we'll start with an easy exercise,
for other rules might have to be known, such as the chain rule, for a complicated exercise.

Example:

y = (x^{3} + 1)(2x^{2} + 8x - 5)

(dy/dx)

= (x^{3} + 1)(dy/dx)(2x^{2} + 8x - 5)
+ (2x^{2} + 8x - 5)(dy/dx)(x^{3} + 1)

= (x^{3} + 1)(4x + 8) + (2x^{2} + 8x - 5)(3x^{2})

= (4x^{4} + 8x^{3} + 4x + 8) + (6x^{4} + 24x^{3} - 15x^{2})

= 10x^{4} + 32x^{3} - 15x^{2} + 4x + 8

This answer could also be found if y was multiplied out and then derived, which is a good way to check the answers of these exercises.