Conference rusure::math

34.0. "Another view of calculus" by RANI::LEICHTERJ () Tue Feb 14 1984 00:53

The usual presentations of calculus give you a bunch of rules for calculating
derivatives; for example, you learn the sum rule, the product rule, and so on.
There is another rule, which I first observed years ago, and a friend finally
proved valid, and christned the "diagonal rule":  Given an expression involving
x in several places, to take the derivative with respect to x, go through all
the places x occurs as a variable one after another, and consider that single
x as a variable, and the rest of the x's to be constants.  Differentiate.
Now add up all the terms you've gotten.

For example, you can get the product rule directly from the diagonal rule:

	d/dx (f(x)g(x)) = f'(x) g(x)	-- first x a variable, second a constant
			+ f(x) g'(x)	--second x a variable, first a constant

In fact, it is easy to show that all the usual rules can be derived from the
three rules:

a) diagonal rule;
b) chain rule;
c) d/dx (1/x) = -1/x^2

Problem:  Prove the diagonal rule is correct.
Hint:  Why is it called that?
							-- Jerry

Jerry,

I would say what you are doing is expressing  Y= f(X)  as  Y= f(X1,X2,...,Xn)

and calculating the partial derivatives of Y with respect to the various Xi.

Then you are using the formula:

              ---
	dY    \    6Y    dXi
	-- =   >   --- * ---
	dX    /    6Xi   dX
              ---

(For proof of this formula, see your favorite Advanced Calc. book.)

                     dXi
But  Xi = X  and so  --- = 1 and your result follows.
                     dX

Mike.

Yup, that's it - it's the diagonal rule because, looking at it closely, you
see that you are replacing f(x) by F(DELTA(x)), where F is f with each x
replaced by a unique xi, i = 1 ... n, and DELTA(x)=(x,x,...,), i.e. embedding
into the diagonal.  The "rule" follows from the chain rule and the total
derivative formula.
							-- Jerry