The $n$-th derivative of the product of three functions

Question

I would like to ask what are the general formula for the the $n$-th derivative of the product of three functions?

For the product of two functions, we have the Leibniz's rule for the general formula, how about three functions?

Example may be like$\ y=\ x^2\cdot x\cdot \ln x$

Answer 1

My answer to your original question is the following: The general form will follow a multinomial pattern. Let $u,v,w$ be functions of $x$. Then:

$\frac{d^{n}}{dx^{n}}(uvw)=\sum_{k_{1}+k_{2}+k_{3}=n}\binom{n}{k_{1},k_{2},k_{3}}u^{(k_{1})}v^{(k_{2})}w^{(k_{3})}$

where $u^{(k_{1})}$, $v^{(k_{2})}$, and $w^{(k_{3})}$ denote the $k_{1}^{th}$, $k_{2}^{th}$, and $k_{3}^{th}$ derivatives of $u$, $v$, and $w$ respectively.

Even more generally if we have a product of $m$ functions $u_{1},...,u_{m}$ of $x$ then:

$\frac{d^{n}}{dx^{n}}(u_{1}\cdots u_{m})=\sum_{k_{1}+...+k_{m}=n}\binom{n}{k_{1},...,k_{m}}u_{1}^{(k_{1})}\cdots u_{m}^{(k_{m})}$.

Your example $f(x)=x^{2}\cdot x\cdot\ln(x)=x^{3}\ln(x)$ so $n^{th}$ derivatives can be handled by the Leibniz formula.

My answer to your new question is the following: The derivatives of $\sin(x)$ and $\cos(x)$ "cycle" so the general pattern is easy to find.

Note that:

$\frac{d}{dx}(\sin(x))=\cos(x)$

$\frac{d}{dx}(\cos(x))=-\sin(x)$

$\frac{d}{dx}(-\sin(x))=-\cos(x)$

$\frac{d}{dx}(-\cos(x))=\sin(x)$.

So $\frac{d^{4n}}{dx^{4n}}(\sin(x))=\sin(x)$

$\frac{d^{4n+1}}{dx^{4n+1}}(\sin(x))=\cos(x)$

$\frac{d^{4n+2}}{dx^{4n+2}}(\sin(x))=-\sin(x)$

$\frac{d^{4n+3}}{dx^{4n+3}}(\sin(x))=-\cos(x)$

A similar pattern holds for $\cos(x)$.

The derivatives of $\tan(x)$, $\cot(x)$, $\sec(x)$, and $\csc(x)$ don't behave as nicely. Apparently it is possible to get a general formula for the $n^{th}$ derivative of $\tan(x)$ and $\cot(x)$ as seen in http://arxiv.org/pdf/1202.1205.pdf . Unfortunately I don't know of a general pattern for $\sec(x)$ and $\csc(x)$ so I give you the following:

For $\sec(x)$ we have that

$\frac{d}{dx}(\sec(x))=\sec(x)\tan(x)$.

Letting $f(x)=\sec(x)$ this tells us that (assuming $\tan(x)>0$, a similar formula will hold for $\tan(x)<0$) $f'(x)=\sec(x)\tan(x)=\sec(x)\sqrt{\sec^{2}(x)-1}=f(x)\sqrt{f^{2}(x)-1}$. So denoting the $n^{th}$ derivative of $f$ by $f^{(n)}(x)$ we have:

$\frac{d^{n+1}}{dx^{n+1}}(\sec(x))=f^{(n+1)}(x)=(f')^{(n)}(x)=(f\sqrt{f^{2}-1})^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}f^{(n-k)}(x)(\sqrt{f^{2}(x)-1})^{(k)}=\sum_{k=0}^{n}\binom{n}{k}(\sec(x))^{(n-k)}(\sqrt{sec^{2}(x)-1})^{k}$.

You can expand out the term involving the square root using Faà di Bruno's formula (http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno's_formula).

These formulas give the general derivatives of $\sec(x)$ in terms of lower order derivatives. Similar techniques can be used to calculate the general form of the derivatives of $\csc(x)$. I don't know the exact general pattern for these functions though.