The Unapologetic Mathematician

Mathematics for the interested outsider

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


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


Now we can find the differentials of p and q


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


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}

October 9, 2009 - Posted by | Analysis, Calculus

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