The method employed by Lagrange may be used for the determination of the successive differentials of other functions. (25) Let u= €o12. If a become x+h, u becomes (+)2 = = © (x2 + 2x h+h2) Cz2 Multiplying these together, taking only the coefficient of h' and multiplying it by 1.2...r, we find (26) From this we can determine the successive differentials of cos 2 and sin x2. Let u = cos x2 + ( − )1 sin a2 (−) sin x = g(-)* Then differentiating by the preceding formula ďu = c(-)* a2 { (−)13 (2x)' + (−)'=' r(r − 1) (2.x)r-2 dx = and c(-)*2 (-)*p! = COS (2x)'1+ &c.}. (x2 + p27) + (−)3 sin (x2 + p Therefore making these substitutions, and as π d' (cos a') = (2.8)' cos(22+r) +r(r−1)(2x)'=' cos{s' + (r− 1)"} dx d' (sin x2) dx = (2a)′ sin (a2+r7") +r(r−1) (2a)′-2 sin{x2 + (r − 1)7} r (r − 1)...... († − 3) (2 a)*-* sin {x2 + (r − 2) T} 1.2 + &c. We might in this case expand the function and differentiater times each term in the development, but as this would give d'u dx expressed in an infinite series, the following method, due to Laplace*, is to be preferred. It is easily seen on effecting two or three differentiations that the form Hence multiplying by (e* + 1)*+1 we must have The product of (2) and (3) must be equal to the second side of (1), and as this last consists of a finite number of terms having positive indices, the terms in the product of (2) and (3) which contain negative indices must disappear of themselves. Hence taking the terms with positive indices |