be simultaneously zero, we must proceed to second differentials, an expression generally indeterminate by reason of the indefiniteness of the ratio of dx to dy. If all the partial differentials of f and F of the second order are zero, we must proceed to the third order, and so on. The extension of the preceding considerations to indeterminate functions of any number of independent variables is obvious. We have considered only the case of indetermination the application of the method, however, to that of the form of the form 0 ∞ ∞ may be established just as in the instance where a is an arbitrary quantity. Thus p(x, y) may have any 3 hence (x, y) = -, a determinate value. == The partial differentials of the first order being zero, we must proceed to differentials of the second order. a being an arbitrary quantity. The value of p(x, y) is therefore indeterminate, within certain limits; its greatest and least values corresponding to the least positive and least negative values of a + then 1. Suppose that a hence + 2 and 2 are the least positive and negative values of 1 B or a + a It appears therefore that p(x, y) may have any value whatever from 0 to + 2. Evaluation of indeterminate implicit Functions of a single Variable. 67. Suppose that an equation f(x, y) = 0.. .(1) is satisfied identically by a certain value x, of x, whatever be the value of y. The function y will for this value of x appear to be indeterminate. Differentiating the proposed equation, we get 0 But since, when x = x ̧, ƒ(x, y) has a constant value zero for all values of y whatever, it follows that in this case also The value of y, must be determined from the equation (3). In case the equation (3) be satisfied identically for all values of y, we must, the function df dx now occupying the place of the original function f, proceed to determine y, from the equation and so on, until the indeterminateness is eradicated. ƒ (x, y) = (y2 − 1) x2 − y {log (1 + x)}2 = 0, x = 0. or F(x, y) = (y2 − 1) (x2 + x) − y log (1 + x) = 0 : but this equation is identically satisfied by x = 0: we must therefore differentiate again with regard to x. CHAPTER VI. MAXIMA AND MINIMA. Definition of a Maximum and Minimum. 68. LET y = f(x), and suppose that, as a gradually increases, through a particular value x, from a value x-h to a value x+h, h being an indefinitely small positive quantity, y increases, as ≈ increases from x ̧ h to x, and decreases, as x increases from x。 to x + h. In this case y is said to have a maximum value when x = x. If y decreases, as x increases from x-h to x, and increases, as x increases from x, to x+h, then y is said to have a minimum value when x = xo. The words increase and decrease are here used in algebraical senses to indicate progress from towards, and from + ∞ o 69. Before proceeding to investigate a rule for determining the maximum or minimum values of y, it will be necessary to premise the following lemma. LEMMA. If u be a function of x; then, accordingly as u du is increasing or decreasing as x increases, tively positive or negative, and conversely. dx will be respec Suppose that, when x increases to a value x + dx, dx being very small, u becomes u+ du; then, if u be increasing with the increase of x, Su must be positive, and if u be decreasing, Su must be negative: hence Su must be positive in the former |
