# The Unapologetic Mathematician

## Product and Quotient rules

As I said before, there’s generally no product of higher-dimensional vectors, and so there’s no generalization of the product rule. But we can multiply and divide real-valued functions of more than one variable. Finding the differential of such a product or quotient function is a nice little exercise in using Cauchy’s invariant rule.

For all that follows we’re considering two real-valued functions of $n$ real variables: $f(x)=f(x^1,\dots,x^n)$ and $g(x)=g(x^1,\dots,x^n)$. We’ll put them together to give a single map from $\mathbb{R}^n$ to $\mathbb{R}^2$ by picking orthonormal coordinates $u$ and $v$ on the latter space and defining

$\displaystyle\begin{aligned}u&=f(x)\\v&=g(x)\end{aligned}$

We also have two familiar functions that we don’t often think of explicitly as functions from $\mathbb{R}^2$ to $\mathbb{R}$:

$\displaystyle\begin{aligned}p(u,v)&=uv\\q(u,v)&=\frac{u}{v}\end{aligned}$

Now we can find the differentials of $p$ and $q$

$\displaystyle\begin{aligned}dp(u,v)&=vdu+udv\\dq(u,v)&=\frac{1}{v}du-\frac{u}{v^2}dv=\frac{vdu-udv}{v^2}\end{aligned}$

Notice that the differential for $q$ is exactly the alternate notation I mentioned when defining the one-variable quotient rule!

With all this preparation out of the way, the product function $f(x)g(x)$ can be seen as the composition $p(f(x),g(x))$, while the quotient function $\frac{f(x)}{g(x)}$ can be seen as the quotient $q(f(x),g(x))$. So to calculate the differentials of the product and quotient we can use Cauchy’s invariant rule to make the substitutions

$\displaystyle\begin{aligned}u&=f(x)\\v&=g(x)\\du&=df(x)\\dv&=dg(x)\end{aligned}$

The upshot is that just like in the case of one variable we can differentiate a product of two functions by differentiating each of the functions, multiplying by the other function, and adding the two resulting terms. We just use the differential instead of the derivative.

$\displaystyle d\left[fg\right](x)=df(x)g(x)+f(x)dg(x)$

Similarly, we can differentiate the quotient of two functions just as in the one-variable case, but using the differential instead of the derivative.

$\displaystyle d\left[\frac{f}{g}\right](x)=\frac{df(x)g(x)-f(x)dg(x)}{g(x)^2}$

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Posted by John Armstrong | Analysis, Calculus

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