# Implicit Differentiation… the triple product rule?

When implicitly finding the derivative of:

$xy^3 - xy^3\sin(x) = 1$

How do you find the implicit derivative of:

$xy^3\sin(x)$

Is it using a triple product rule of sorts?

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Using the product rule twice tells you what the triple product rule looks like. –  Qiaochu Yuan Sep 1 '10 at 2:57
Can you give an example? –  KronoS Sep 1 '10 at 3:04
@KronoS: Why don't you try computing the derivative of x^3 and then for xxx using the "triple product rule". If you've got the right rule, you'll get the same answer. –  Jason DeVito Sep 1 '10 at 3:08
@Kronos Consider three functions f, g, and h and consider the triple product f x ( g x h ) and we want to take the derivative of this if they are all functions of variable X. Then the first product rule tells us to take the derivative of f and the derivative is what is in parenthesis, now think about what the derivative of what is in parenthesis. –  BBischof Sep 1 '10 at 3:08
@BBischof I GET IT!!! sweet thanks! if you answer it I'll select it for you. –  KronoS Sep 1 '10 at 3:10

HINT $\;$ logarithmic differentiation makes the n-ary generalization obvious:

$$\rm (abc)'= abc \; log(abc)' = abc \;(log\; a + log\; b + log\; c)' = abc \; \bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c}\bigg)$$

Obviously the same proof works for arbitrary length products yielding

$$\rm (abc\cdots f)' = \: abc\cdots f\;\bigg(\frac{a'}{a} + \frac{b'}{b} + \frac{c'}{c} +\:\cdots\:+ \frac{f'}{f}\bigg)$$

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It is possible to derive a 'triple product rule.' In fact, you can generalize it to an arbitrary number of products. You may ask why these rules aren't presented in these general forms. One reason for this is that writing the general formula out is a more complicated to memorize and more intimidating towards students just learning calculus. The other reason, is that having the product rule is enough to give you the general rule by a simple induction argument. Also, for practical purposes, problems can be solved with only the usual product rule being applied in multiple steps.

In your case we can apply the product rule twice: $\frac{d}{dx}xsin(x)y^{3}=y^{3}\frac{d}{dx}(xsin(x))+3y^2\frac{dy}{dx}(xsin(x))=y^{3}(sin(x)+xcos(x))+3y^{2}\frac{dy}{dx}$

We can also derive a 'triple product rule'

$\frac{d}{dx}(f\cdot g\cdot h)=\frac{df}{dx}\cdot(h\cdot g)+\frac{dg}{dx}\cdot (f\cdot h)+\frac{dh}{dx}\cdot (f\cdot g)$

But you see that you still have to apply the chain rule to that to get it to fit how you need to use it in this situation and so it would be a real mess to memorize all the different special cases. This is why we just provide the product rule with 2 functions.

Does my rambling make sense?

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I don't see that "This is why we just provide the product rule with 2 functions." Only the slightest increase in mental energy is required to learn a product rule with 3 (or 4 or "n") terms. Instead of the standard "(derivative of the 1st, times the 2nd), plus (derivative of the 2nd, times the 1st)", one merely thinks "(derivative of the 1st, times everything else), plus (derivative of the 2nd, times everything else), plus (derivative of the 3rd, times everything else), ...". This generalization is handy to know, and is a reason not to "just provide the product rule with 2 functions". –  Blue Sep 1 '10 at 4:03
I made that statement because its easy to see how to get the general rule from the rule for 2 cases and presenting a 'generalized' product rule for n-products would probably just be intimidating for many students in a first calculus course. I agree that the general rule is useful to understand, but I was trying to say that there is no point in simply memorizing a general formula because the important piece is understanding how it extrapolates to an arbitrary number of variables. In my view, its similar to presenting commutativity with only 2 elements, even though it generalizes to n elements. –  WWright Sep 1 '10 at 19:57

Here is a simple way to compute the $n$-th derivative of a product of $k$ functions. To ease the notation, stick to the case $k=3$. First follow your instinct, and then you'll see that the justification presents no difficulty. Write (with all summation indices varying from 0 to $\infty$) $$\sum_n\frac{(fgh)^{(n)}}{n!}X^n =\sum_i\frac{f^{(i)}}{i!}X^i\ \sum_j\frac{g^{(j)}}{j!}X^j\ \sum_k\frac{h^{(k)}}{k!}X^k$$ $$=\sum_{i,j,k}\frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}X^{i+j+k} =\sum_n\ X^n\sum_{i+j+k=n}\frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}\quad.$$

In categorical language: To each $\mathbb Q$-algebra $A$ is attached the $\mathbb Q$-algebra $A':=A[[X]]$ equipped with the derivation $d/dX$ (the letter $X$ being an indeterminate). Then $A\mapsto A'$ is a right adjoint to the forgetful functor from differential $\mathbb Q$-algebras to $\mathbb Q$-algebras.

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Readers may find it helpful to first see my remark here on this generating-function approach. –  Bill Dubuque Sep 1 '10 at 16:36

REMARK $\$ Pierre's answer is a generating function approach:

$\displaystyle\rm\quad\quad\quad\;\; f(x+t) \;=\; \sum_{n=0}^\infty \;\; f^{(n)}(x) \; \frac{t^n}{n!}$

$\rm\quad (fgh)(x+t) \;=\; f(x+t)\: g(x+t)\: h(x+t)$

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\;\: \;=\; (f + f' t +\:\cdots) (g + g' t + \:\cdots) (h + h' t + \:\cdots)$

$\rm\quad\quad\quad\quad\quad\quad\quad\quad\;\: \;=\; fgh + (f'gh + fg'h + fgh')\: t + \:\cdots$

$\rm\quad (fgh)(x+t) \;=\; fgh + (fgh)' t + \:\cdots$

Thus we conclude $\:\rm(fgh)' = f'gh + fg'h + fgh'$

This generalizes to n-th order derivatives, as Pierre mentioned, e.g.

$\quad\displaystyle\rm\sum_{n=0}^\infty \: (fgh)^{(n)} \frac{t^n}{n!} \ =\ \sum_{n=0}^\infty \:\bigg(\ \sum_{i+j+k=n} \frac{f^{(i)}g^{(j)}h^{(k)}}{i!\ j!\ k!}\bigg) \frac{t^n}{n!}$

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If I understand (at least in part) your answer math.stackexchange.com/questions/3628/…, this is a very particular case of umbral calculus. I'm wondering if the adjunction statement in my answer is also a very particular case of something. Any idea? –  Pierre-Yves Gaillard Sep 1 '10 at 17:10

You get a formula for the derivative of a product of $n$ factors from the formula for the product of $2$ factors by doing induction. Intuitively, you do it the same way as you go from $2$ to $3$: $(fgh)' = ((fg)h)' = (fg)'h + (fg)h' = (f'g + fg')h + (fg)h' = f'gh + fg'h + fgh'$. The pattern should now be clear.

Assuming you already know that $(f_1\cdots f_k)' = f_1'f_2\cdots f_k + f_1f_2'f_3\cdots f_k + \cdots + f_1\cdots f_{k-1}f_k'$, then inductively you have $$(f_1\cdots f_kf_{k+1})' = (f_1\cdots f_k)'f_{k+1} + (f_1\cdots f_k)f_{k+1}'$$ and the formula follows.

So for $xy^3\sin(x)$, the derivative will be $$(x)'y^3\sin(x) + x(y^3)'\sin(x) + xy^3(\sin x)'$$ with the middle summand requiring implicit differentiation.

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