Next: The Quotient Rule Up: Derivatives: How to Find Previous: The Power Rule

  
The Product Rule

Okay, here is a key rule, a rule without which calculus would just be a case of goofing around with a pile of very simple functions. Suppose that we have a function h which is obtained by multiplying together two functions f and g. So h(x) = f(x) g(x). And silly us, we want to take the derivative of h. It's extremely tempting to say: $ \displaystyle \frac {d} {dx} (fg) = f ' g ' $, but unfortunately that would be WRONG!


How do we know it's wrong? Well, we didn't waste the best years of our lives studying math for nothing. Besides, if we apply it when f(x) = x and g(x) = x, we get $\displaystyle \frac {d} {dx} (x)(x) = (1)(1) = 1.$ WHICH IS WRONG! We already know that

\begin{displaymath}\frac {d} {dx} (x)(x) = \frac {d} {dx} x^2 = 2x,
\end{displaymath}

not the same as 1.

Instead, the correct answer is found by the following, the real, correct, How to Ace Calculus certified PRODUCT RULE :

\begin{displaymath}\boxed{\frac {d} {dx} (fg) = f'g + fg'}\end{displaymath}

It looks a bit messy, but actually it's extremely simple. To say it in words:

``The derivative of the product of two functions is the derivative of the first times the second plus the first times the derivative of the second."

Let's apply it to the same case we looked at before, when f(x)=x and g(x)=x.

Then the rule says:


\begin{displaymath}\frac {d} {dx} (x)(x) = (x)'(x)+(x)(x)'=(1)(x)+(x)(1)=2x
\end{displaymath}

That is exactly what the power rule says that the derivative of x2 should be. So at least in this one case, the rules are consistent. And consistency is what pudding is all about. And mathematics too.



Next: The Quotient Rule Up: Derivatives: How to Find Previous: The Power Rule
Joel Hass
1999-05-26