By A. F. Bermant

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**Additional resources for A Course of Mathematical Analysis, Part II **

**Example text**

Example 1. We take the functIOn r = yx2 + y~ (the radius vee· tor of th' point P(x, y)). sitive direction of Ox (see Sec. ~14), we have I--~" ;~ = r~ cos LX + r; sin LX = cos cp cos LX + ~in cp sin LX = cos (cp -LX). In particular, the derivative 01 the radius vector with respect to its own direction (LX = cp) is always equal to unity, whilst it is always zero with respect to the perpendicular direction. This has a simple mean· ing: the radius vector changes uniformly with respect to its own direction, with a rate equal to unity, whilst it does not change at all in a direction perpendicular to it.

W) = du + dv + ... [cp(u, v, ... , w)] = f'[cp(u, v, ... , w)] dcp(u, v, ... , w). These formulae follow at once from the property of invariance of the form of the first differential. We shall prove the second formula for illustration. Since the form of the differential does not depend on the nature of the arguments, we assume that these are independent variables. Then d(uv) a (uv) = a;; dv a (uv) du = u dv + v duo + au This is what we wanted to show. The remaining rules are proved in a similar manner.

Tt;(xo, Yo) sin a. f~ (xo, Yo) =f~(xo' Yo) cos a Proof. l of higher order than f(x o + e cos 0;, y() + e sin 0;) --= f(;<:o, e = f~(:t:o, + s, Yo) cos 0; (2. Hence Yo) + f~(xo, Yo) sin + !... e 0; Since sje -+ 0 as e -» 0, the limit of the ratio on the left-hand side exists and is equal to f~(~;o, Yo) cos 0; + f~(xo' Yo) sin 0;. Consequently, given the differentiability of z = f(x, y) at Po(xo, Yo)' we have f~ (x o' Yo) = f~ (xo' Yo) cos 0; + f~ (xo, Yo) sin 0;. This is what we had to prove. If the point Po (xo, Yo) is fixed, the derivative with respect to direction 0; is a function of 0; only (0";; 0; < 2n).

### A Course of Mathematical Analysis, Part II by A. F. Bermant

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