Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have the following equation, I know I can differentiate it using the quotient rule, by first factoring out $\frac{3}{2}$:

$$f(w)=\frac{3(60-w)(w-2)}{2w}$$

But is there a way to differentiate it using the product rule or some kind of the chain rule by looking at it like so:

$$f(w)=3(60-w)(w-2)(2w)^{-1}$$

I guess what I'm asking is, I know how to use the product rule when I have 2 terms, but how do I do it with 3+? Do I derive the first two terms, then use that derived answer with the product rule with the remaining term?

Thanks!

share|improve this question
2  
$\text{quotient rule} = \text{product rule} + \text{chain rule}$ –  cardinal Oct 23 '11 at 23:10
2  
$(fgh)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h'$. –  cardinal Oct 23 '11 at 23:12

4 Answers 4

up vote 4 down vote accepted

The product rule for three terms can be deduced from the product rule for two: $$(fgh)' = \Bigl( (fg)h\Bigr)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h' = f'gh + fg'h + fgh'.$$ That is, it's just "pass-the-prime".

This holds for any number of factors, and can be established by induction. Assuming that we know that $$\Bigl(f_1\cdots f_n\Bigr)' = f_1'f_2\cdots f_n + f_1f_2'f_3\cdots f_n + \cdots + f_1\cdots f_{n-1}f_n',$$ then $$\small\begin{align*} \Bigl(f_1\cdots f_nf_{n+1}\Bigr)' &= \Bigl( (f_1\cdots f_n)f_{n+1}\Bigr)'\\ &= (f_1\cdots f_n)'f_{n+1} + (f_1\cdots f_n)f_{n+1}' \\ &= \Bigl(f_1'f_2\cdots f_n + f_1f_2'f_3\cdots f_n + \cdots + f_1\cdots f_{n-1}f_n'\Bigr)f_{n+1} + (f_1\cdots f_n)f_{n+1}'\\ &= f_1'f_2\cdots f_nf_{n+1} + f_1f_2'f_3\cdots f_nf_{n+1} + \cdots + f_1\cdots f_n'f_{n+1} + f_1\cdots f_nf_{n+1}'. \end{align*}$$

share|improve this answer
    
Great, thank you so much! –  Josh Oct 23 '11 at 23:11
    
So is it suggested to just use the quotient rule in this case? Or is it ok to just use the product rule. That is, which will be the more efficient way, in general, and for this problem? –  Josh Oct 23 '11 at 23:15
1  
@Josh: Whichever way you want it: you will either use the product rule for three factors and the chain rule, or the quotient rule and the product rule for two factors. Personally, for this problem, I would use algebra to rewrite it as$$\frac{3}{2}(60-w)\left(1 - 2w^{-1}\right)$$ and just use the standard product rule. –  Arturo Magidin Oct 23 '11 at 23:17

The answer to your last question is ‘yes’: you can differentiate $uvw$ by treating it as $(uv)w$. In fact you can do this in general to discover an extended form of the product rule: $$(uvw)' = (uv)w'+(uv)'w = uvw' + (uv'+u'v)w = uvw' + uv'w + u'vw\;.$$ (Here I’m thinking of $u,v$, and $w$ as functions of the same independent variable, say $x$.) The same idea extends to more factors. For instance, $$(uvwz)' = (uvw)z'+(uvw)'z = uvwz'+uvw'z + uv'wz + u'vwz\;,$$ using the previous result.

share|improve this answer

The general product rule for more than two terms is to differentiate one term at a time. For example, the derivative of a product of three functions: $$ \frac{d}{dx}(f g h) = \frac{df}{dx}gh+f\frac{dg}{dx}h+fg\frac{dh}{dx} $$ This is, of course, the same as applying the product rule for two terms twice as you suggested: $$ \frac{d}{dx}((f g) h) = \left(\frac{df}{dx}g+f\frac{dg}{dx}\right)h+(fg)\frac{dh}{dx} $$

share|improve this answer

To answer the second question first: You can certainly apply the product rule to each multiplication in turn. However, since multiplication is commutative and associative, the result will necessarily be symmetric with respect to the three factors, so it makes sense to derive a general rule and remember it instead of going through the factors pairwise in each case.

Applying the product rule to the product $fgh$ by grouping it as $(fg)h$ yields

$$(fgh)'=((fg)h)'= (fg)'h + (fg)h' = (f'g + fg')h + (fg)h'=f'gh+fg'h+fgh'\;.$$

Perhaps you can guess from this result what the general result for an arbitrary number of factors will be:

$$\left(\prod_i f_i\right)'=\sum_if_i'\prod_{j\ne i} f_j\;.$$

You can either prove this using induction (which might be a good exercise), or you can see why it's true by considering what happens if you change each of the factors a bit:

$$(f+\Delta f)(g+\Delta g)(h+\Delta h)\cdots=fgh\cdots+\Delta fgh\cdots+f\Delta gh\cdots+fg\Delta h\cdots+\cdots\;,$$

simply from multiplying out the parentheses. The remaining terms are at least of second order (i.e. they contain at least two $\Delta$ terms), so they don't end up in the derivative, which gives the first-order approximation to the function.

Turning now to your first question, yes, you can apply the product rule instead of the quotient rule; well spotted. In fact, the quotient rule is nothing but a convenient shortcut for applying the product rule:

$$\left(\frac fg\right)'=\left(fg^{-1}\right)'=f'g^{-1}+f\left(g^{-1}\right)'=f'g^{-1}+f\left(-\frac{g'}{g^2}\right)=\frac{f'g-fg'}{g^2}\;.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.