Implicit Differentiation... the triple product rule?

Question

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?

Answer 1

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?

Answer 2

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

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

Answer 3

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.

Answer 4

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.

Answer 5

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!} $