# Using the product rule to show that $(fgh)' = f 'gh + fg' h + fgh'$

I understand the product rule, as conventionally stated:

$$(fg)' = f 'g + f g'.$$

But the above only involves two functions ($f$ and $g$), whereas the question I'm asking involves 3. How would I go about proving this? El VBbvBN

Initial note: I can see you are using the product rule when only three functions are involved, but why stop there? Why not generalize it? Generalizing it will solve your specific problem and also when you would have $4, 5, \ldots, n$ functions. Doing so requires a more sophisticated proof, but induction will take care of it fairly easily. Thus, I will provide a quick solution to your specific problem (similar to the idea encouraged by mm-aops but with the details filled in), and then I will give you a proof of the generalized product rule.

Specific case: You want to prove that $$(fgh)' = f'gh + fg'h + fgh',\tag{1}$$ where you are are allowed to use the "basic product rule" $$(fg)' = f'g + fg'.\tag{2}$$ To this end, let $\eta=fg$. Then \begin{align} (\eta h)' &= \eta'h + \eta h'\tag{used $(2)$}\\[0.5em] &= (fg)'h + (fg)h'\tag{since $\eta=fg$}\\[0.5em] &= (f'g+fg')h + fgh'\tag{used $(2)$ again}\\[0.5em] &= f'gh+fg'h+fgh'.\tag{expanded} \end{align} Thus, your specific problem is done. $\Box$

Generalized product rule: We are going to conjecture that $$(f_1f_2\ldots f_n)'(x)=f'_1(x)f_2(x)\ldots f_n(x)+f_1(x)f'_2(x)\ldots f_n(x)+\cdots+f_1(x)f_2(x)\ldots f'_n(x).$$ Suppose the statement $P(n)$ is $$P(n) : (f_1f_2\ldots f_n)'(x) = f'_1f_2\ldots f_n+f_1f'_2f_3\ldots f_n+\cdots+f_1f_2\ldots f'_n.$$ Observe that $P(1)$ is trivially true. Now, suppose that $P(k)$ is true for some integer $k$; that is, $$(f_1f_2\ldots f_k)'(x) = f'_1f_2\ldots f_k+f_1f'_2f_3\ldots f_k+\cdots+f_1f_2\ldots f'_k.$$ We want to show that $P(k+1)$ is true; that is, $$(f_1f_2\ldots f_{k+1})'(x) = f'_1f_2\ldots f_kf_{k+1}+f_1f'_2f_3\ldots f_kf_{k+1}+\cdots+f_1f_2\ldots f_kf'_{k+1}.$$ We use $(2)$ to write the following: \begin{align} (f_1f_2\ldots f_{k+1})'(x) &= [(f_1f_2\ldots f_k)f_{k+1}]'(x)\\[0.5em] &= (f_1f_2\ldots f_k)f'_{k+1}+f_{k+1}(f_1f_2\ldots f_k)' \\[0.5em] &= (f_1f_2\ldots f_k)f'_{k+1}+f_{k+1}[(f'_1f_2\ldots f_k)+(f_1f'_2\ldots f_k)+\cdots+(f_1f_2\ldots f'_k)]\\[0.5em] &= f'_1f_2\ldots f_kf_{k+1}+f_1f'_2f_3\ldots f_kf_{k+1}+\cdots+f_1f_2\ldots f'_kf_{k+1}+f_1f_2\ldots f_kf'_{k+1} \end{align} This is exactly what we wanted to show. Thus, by mathematical induction, $P(n)$ is true for all $n\in\mathbb{N}$. $\Box$

Note: The result just established confirms your specific problem for when $n=3$. To see the value in establishing the general result, suppose you had to consider the case where $n=4$; that is, what would $(fghj)'$ be? By the generalized product rule just proven, you can simply state that $$(fghj)' = f'ghj + fg'hj + fgh'j + fghj'$$ without going into messy algebraic manipulations like all of the answers here have done (including my answer to your specific problem).

• @H_T Thanks, glad you found it helpful. :) – Daniel W. Farlow Mar 14 '15 at 16:45

Define the function $m := fg$. Then $fgh = mh$. use the product rule for $(mh)'$ and then use it again to compute $m'$.

here is one more way to this without the product rule at all. taking $\ln$ of $$y = fgh$$ gives you $$\ln y = \ln f + \ln g + \ln h$$ taking the derivative gives you $$\frac 1y\frac{dy}{dx} = \frac {f'}f + \frac {g'}g + \frac {h'}h$$ cleaning it up gives you $$(fgh)' = {f'}gh + f {g'}h + fg {h'}$$

• That was cute. (+1) – Daniel W. Farlow Feb 23 '15 at 2:40

Hint: Use the product rule twice. First with $$(f \cdot (gh) )' = f' \cdot (gh) + f \cdot (gh)'.$$ You work out the rest.