Course:MATH110/Archive/2010-2011/003/Notes/Product Rule and Quotient Rule

This note describes two important rules to take derivatives of more complex functions.

The Product Rule

Let $f$ and $g$ be two functions for which we know the derivative. What is the derivative of their product?

Recall: Product of two functions
Recall that if $f$ and $g$ are two functions, their product is the new function $fg$ which is defined as follow: $(fg)(x)=f(x)g(x)$ In other words, the value of the function $fg$ at the point $x$ equals the product of $f(x)$ and $g(x)$.

The answer to the above question is given by the product rule which is stated below:

Product Rule
Let $f$ and $g$ be two functions and denote by $h=fg$ their product, then $h'(x)=(fg)'(x)=f'(x)g(x)+f(x)g'(x)$

It is important for you to try by yourself on examples to see why this rule is true and not the more simplistic rule that one could imagine of having the derivative of the product of two functions to simply be the product of its respective derivatives. Try for yourself, pick two functions for which you know the derivative, but for which you also already know the derivative of the product and then check for yourself.

Then, practice the product rule as much as possible. Problems 7 to 16 in section 3.3 of the textbook is a good start for your practice.

The Quotient Rule

Let $f$ and $g$ be two functions for which we know the derivative. What is the derivative of their quotient?

Recall: Quotient of two functions
Recall that if $f$ and $g$ are two functions, their quotient is the new function ${\frac {f}{g}}$ which is defined as follow: $\left({\frac {f}{g}}\right)(x)={\frac {f(x)}{g(x)}}$ In other words, the value of the function ${\frac {f}{g}}$ at the point $x$ equals the quotient of $f(x)$ by $g(x)$.

The answer to the above question is given by the quotient rule which is stated below:

Product Rule
Let $f$ and $g$ be two functions and denote by $h={\frac {f}{g}}$ their quotient, then $h'(x)=\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)g(x)-f(x)g'(x)}{(g(x))^{2}}}$