Product rule for differentiation is different from the sum and difference rule for derivatives. In sum and difference rules the derivative of a sum or difference is the sum or difference of the derivatives. In other words we find the derivatives separate and add or subtract as required.
But the derivative of the product of a function is not the product of derivatives.
For example, consider f(x) = x5. By using the power rule for derivative of a function we get f'(x) = 5x4.  If we treat the function f(x) as product as f(x) = x3.x2, then  the derivative of x3 = 3x2 and the derivative of x2 = 2x. If we multiply the two derivatives we get (3x2).(2x) = 6x3 which is not the derivative of x5 .Hence the differentiation of a product is analyzed using the definition of the derivative and a rule is arrived at for finding the derivative of a function which is expressed as a product of two or more functions.
Many rules exist under the name Product Rule related to various concepts like the product rule for exponents, product rule for logarithms, product rule for limits and product rule for derivatives. Let us first view few instances where the product rules are defined other than for derivatives.

The Product Rule for derivatives can be defined as follows.

If two functions are differentiable, so is their product and the derivative of their product is the sum of the products obtained by multiplying a function with the derivative of the other function.

Using notation we can write this as:

$\frac{d}{dx}$ $[f(x)g(x)]$ = $f(x)g'(x)+g(x)f'(x)$
Product rule for exponents is a basic algebraic rule that is used to simplify when products of same variable with different exponents occur.

$a_m\ x_n$ = a^{m+n}$ where m and n are any real numbers.


$3^4 . 3^3 = 3 ^{4+3} = 3^7$.

$x^7 . x^{-2}$ =$x^{7+(-2)}$ = $x^5$.
Equations involving exponents can also be written as logarithmic equations. Thus we have a counter part of Product rule of exponents for finding the logarithm of a product.

$log_a(xy)$ = $log_a x + log_a y$. where x ≥ 0 and y ≥ 0


$log_{10}$ (10000) = $log_{10}$(10 x 1000) = $log_10$ 10 + $log_10$ 1000

= 1 + $log_{10}$ $10^3$ = 1 + 3 = 4

$log_{a}$1 =$log_a$  ($\frac{1}{a}$) = $log_a (a^1 \times a^{-1})$

= $log_{a}$ a + $log_a a^{-1}$ = 1 + (-1) = 0
The above statement can be stated as the product rule of differentiation as follows.
If u and v are differentiable functions of x, then:

$\frac{d}{dx}(uv)$ = u.$\frac{dv}{dx}$ + v.$\frac{du}{dx}$ The above rule can be proved using the definition of the derivative and the rule for limits.
Just as the derivative of a product is not the product of the derivatives, the integral of a product is also not equal to the product of the integrals. Corresponding the the Product Rule for derivatives, a special technique known as Integration by parts is used to integrate a product.
The formula for integration by parts is as follows:

$\int u\ dv $ = uv - $\int v\ du$

The other version of the same formula is:

$\int f(x)g'(x)dx$ = f(x).g(x) - $\int g(x)f'(x)dx$

The key in finding the integral using integration by parts lies in identifying an easily integrable function as g'(x).

Solved Example

Question: Find $\int xsinxdx$
Let u = x and dv = sin x.
Hence we get du = 1 and v = -cos x
Plugging the above functions in the formula for integration by parts
$\int udv $ = uv - $\int vdu$
$\int xsinxdx$ = x(-cos x) - $\int -cosx.1.dx$
                      = -x cos x + $\int cosxdx$
                      = -x cosx + sin x + c.

When a function is expressed as a product of three factors, then its derivative can be found using triple product rule. The triple product rule is derived from the product rule for two factors, by clubbing two factors together as a single factor.
$\frac{d}{dx}$ (uvw) = u'vw + v'wu + w'uv where u, v and w represent the functions of x.
If p(x) = f(x).g(x).h(x), then

p'(x) = f'(x).g(x).h(x) + g'(x).h(x).f(x) + h'(x).f(x).g(x).

Solved Example

Question: Find the derivative of y = x2 ex sec x
Let u(x) = x2, v(x) = ex and w(x) = sec x.
Then u'(x) = 2x, v'(x) = ex and w'(x) = sec x tan x
Using the triple product rule,

$\frac{dy}{dx}$ = u'vw + v'wu + w'uv

                        = 2x.ex.sec x + ex.sec x.(x2) + sec x tan x . x2.ex.
                        = x.ex.sec x(2 + x + x tan x)

The product rule for the derivative can be proved using the limit definition of the derivative and the laws of limits:

The derivative of the product 'uv' where us and v are functions of s is:

 $\frac{d}{dx}$ (uv)= $\lim_{h->0}$ $\frac{u(x+h)v(x+h)-u(x)v(x)}{h}$ Limit definition of the derivative

In order to bring in the derivatives u'(x) and v'(x) in the form of difference quotients we add and subtract u(x).v(x+h) in the numerator.

$\frac{d}{dx}$ $(uv)$= $\lim_{h->0}$ $\frac{u(x+h)v(x+h)-u(x)v(x+h)+u(x)v(x+h)-u(x)v(x)}{h}$

= $\lim_{h->0}$ [$v(x+h)$ $(\frac{u(x+h)-u(x)}{h})$ + u(x) $(\frac{v(x+h)-v(x)}{h}))]$ 
Factored separately.

= $\lim_{h->0}v(x+h)$ . $\lim_{h->0}$ $\frac{u(x+h)-u(x)}{h}$ + $\lim_{h->0}u(x)$ . $\lim_{h->0}$ $\frac{v(x+h)-v(x)}{h}$

= v(x) . u'(x) + u(x) . v'(x) Applied limits and definition of the derivative.
= u(x) . v'(x) + v(x) . u'(x) The Product rule for the derivatives.

The rule can also be proved adding and subtracting u(x + h).v(x) and grouping suitably.

Solved Examples

Question 1: Find the derivative of y = (x +3) (x2 -5)
Let u(x) = x +3 and v(x) = x2 - 5
    Thus u'(x) = 1 and v'(x) = 2x.                      Differentiated using power rule.

    $\frac{dy}{dx}$ = u(x) . v'(x) + v(x) . u'(x)

                            = (x + 3) 2x + (x2 - 5)(1)       Substitution
                            = 2x2 + 6x + x2 - 5
                            = 3x2 + 6x - 5
   The derivative for the above example can also be found without using the product rule, first by expanding the function and applying the power rule separately.

Question 2: Find the derivative of f(x) = ex . tan x
Let u(x) = ex and v(x) = tan x
          u'(x) = ex and v'(x) = sec2 x
    Then applying product rule for differentiation,
    f'(x) = u(x) . v'(x) + v(x) . u'(x)
           = ex sec2 x + ex tan x
           = ex (sec2 x + tan x)
The product rule is particularly useful when the two functions are of different types as in the above example.

   At times the product rule can be used instead of quotient rule to make the differentiation process easier.

Question 3: Find the derivative of = $\frac{ln x}{x^{2}}$
We can find the derivative both by using quotient and product rule. But using product rule will make the simplification easier.
       f(x) = ln x . x-2 .             Used the negative exponent to express the quotient as a product.
       let u(x) = ln x and v(x) = x-2.

       Then u'(x) = $\frac{1}{x}$   and v'(x) = -2x-3 .       Derivatives found using the formulas

       f'(x) = u(x) . v'(x) + v(x) . u'(x)

              = ln x (-2x-3) + x-2 ($\frac{1}{x}$

              = -$\frac{lnx}{2x^{3}}$ + $\frac{1}{x^{3}}$

              = $\frac{2-lnx}{x^{3}}$