ax (18) Let u v = e X, X being any function of x. Then making u = This result, when generalized, is of great importance in the solution of Differential Equations. If the function to be differentiated be (a + bx + cx2)", the general differential might be found by resolving the trinomial a + bx + ca2 into two factors of the first degree, as into (x + a) (x + ẞ), and then differentiating the product (x + a)” (x + ß)" by the Theorem of Leibnitz; but instead of doing so we shall make use of two formulæ given by Lagrange*. Let u= a + bx + c x2, u' = b + 2 c x ; Then substituting x + h for x in u" it becomes Developing it as a binomial, of which u+u'h is the Again, developing each binomial and taking only the terms which multiply h', we find that the term in Collecting these terms, and multiplying by 1.2 ... r, we obtain for the 7th differential coefficient of u" and the 7th differential of u" is the coefficient of h' in this expansion multiplied by 1.2...r. Now expanding each term by the binomial theorem, we have for the coefficient of and so on. Collecting these terms and multiplying by 1.2...r, we find Here u' = 2x, e = 4a2, and if we make r = n, we find by formula (B), d" (a2 + x2)" dx" The 7th differential of this function may be found as in the last example, but the following method gives it under a form which is more convenient in practice; { x − a ( − )}}'+' = (a2 + x2)* {cos (r + 1)9 − (−)1 sin (r + 1) 0}, 2 {x + a(−)}}r+1 = (a2 + x2)* {cos (r + 1)8 + (−)1 sin (r + 1) 0} . |