### Rules of differentiation: Rules of computation for the derivative

### The product rule for differentiation

Product of two functions

Let #f# and #g# be two functions.

The **product** of #f# and #g# is the function that assigns to #x# the value #f(x)\cdot g(x)#. This function is denoted by #f\cdot g#, so the function rule is #(f\cdot g)(x) = f(x)\cdot g(x)#.

For example, if #f(x)=x+1# and #g(x)=x^3+1#, then #f\cdot g# is the function with function rule \[\left(f\cdot g\right)(x)=f(x)\cdot g(x) = \left(x+1\right)\cdot \left(x^3+1\right) = x^4+x^3+x+1\tiny.\]

Product rule for differentiation

The derivative #(f\cdot g)'# of the product #f\cdot g# is given by the productformula

\[(f\cdot g)' = f' \cdot g + f\cdot g'\tiny.\]

This means that #(f\cdot g)'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)# for all #x#.

In order to prove the rule, we rewrite *difference quotient* as follows:

\[\begin{array}{rcl} \text{difference quotient}&=&\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=& \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &=& \frac{(f(x+h)-f(x))g(x)+f(x+h)(g(x+h)-g(x))}{h}\\ &=& \frac{f(x+h)-f(x)}{h}g(x)+f(x+h)\frac{g(x+h)-g(x)}{h}\\ \end{array}\]

Now we can calculate the limit of the difference quotient for #h\to 0#: \[\begin{array}{rcl} \left(f\cdot g\right)'(x) &=& \lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &=&\lim_{h\to 0}\left( \frac{f(x+h)-f(x)}{h}g(x)+f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &=&\lim_{h\to 0}\left(\frac{f(x+h)-f(x)}{h}g(x)\right)+\lim_{h\to 0}\left(f(x+h)\frac{g(x+h)-g(x)}{h}\right)\\ &=&\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\cdot \lim_{h\to 0} g(x)+\lim_{h\to 0}f(x+h)\cdot \lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=& f'(x)\cdot g(x)+f(x)\cdot g'(x)\tiny.\\ \end{array}\]

The product rule gives that the derivative is equal to \[\frac{\dd}{\dd x}(x^2+7\cdot x-3)\cdot (x^3+2\cdot x-1) + (x^2+7\cdot x-3)\cdot\frac{\dd}{\dd x} (x^3+2\cdot x-1)\tiny.\]According to

*The derivative of a polynomial*this is equal to

#(2x+7)\cdot (x^3+2\cdot x-1) + (x^2+7\cdot x-3)\cdot (3x^2+2)#, which can be rewritten as #2\cdot x^4+7\cdot x^3+4\cdot x^2+12\cdot x-7+3\cdot x^4+21\cdot x^3-7\cdot x^2+14\cdot x-6#

and thus is equal to #5\cdot x^4+28\cdot x^3-3\cdot x^2+26\cdot x-13#.

*The derivative of a polynomial*.

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