31.] Differentiation of a compound function, or of a function of a function. Thus far we have calculated the change of value of a function of x, due to a small variation of x; but suppose the function to be compound, and thus to depend on x, not immediately, but by means of some function; what modification must the principles and results of the preceding Articles undergo? The only condition to which a is subject is, that it is a continuous and finite variable for the values assigned to it; and this condition will be fulfilled, if x is replaced by any continuous and finite function of x, and if its variation due to a small variation of x is infinitesimal: in this case therefore the original function will vary in consequence of the variation of this latter function. Pari ratione this latter function may be a compound function, and therefore not vary immediately by reason of the variation of its variable, but only through some other function; and so on to any number of functions. Taking then our former notation, in this case y is a function of a function of x, or a function of many functions of x, and its differential must be calculated as follows: and let us first take the case in which only two functions are involved: viz. where y=f{p(x)}. Since if y = f(x), dy=f'(x) dx; therefore, if x is replaced by p(x), dx must be replaced by the differential of (x), that is, by '(x) dx; and therefore if y = У f{$(x)}, dy = ƒ'{$(x)} $'(x) dx. And so again, if the x involved in p(x) is a function of x, say y(x), (x) will be replaced by {(x)}, and '(x) dx by '{¥(x)} \'(x) dx; and a similar notation will adapt itself to a function of a function...... (to n functions) of x. Thus if y = ƒ{d[¥(x)]}, dy = ƒ'{$[¥(x)]} ¢ ́[4(x)] \'(x) dx. Or we may substitute as follows: But as generally it is convenient to make as few substitutions as possible, the former value of dy is preferable to the latter; the latter also, it is to be observed, is an identity. Hence, by reason of the last Article, if 3(x+3)2 (x2-2)2 dx-4x (x2-2) (x+3)3 dx (x2-2)4 (x+3)2 ( − x2 — 12x-6) (x2-2)3 32.] Differentiation of a*. Let y = a*, dx. Y+▲y = ax+ Ax; ▲y = a¤+sx — aa, = a* {a^*—1}. Let a-1=z, so that az and z are simultaneously infinitesimal; a^x = z+1, ▲x log, a = loge (1 + z), and by Cor. I, Lemma I, log, (1 + z) = z, when z is infinitesimal; dx log, a = 2; and therefore adx-1 must be replaced by dx log, a ; .. dy = a* log, a.dx, d.ax = log, a. a* dx: (8) that is, the differential of a is the product of a*, of the Napierian logarithm of a, and of the differential of the exponent. And hence, if a2 = f(x), ƒ'(x) = log, a.a*. 33.] Differentiation of e*. In the last Article let the base a be replaced by the Napierian PRICE, VOL. I. I base e; in which case log, a becomes log, e, which is equal to 1; and therefore d.ex ex dx: (9) that is, the differential of e* is the product of the quantity and the differential of the exponent. Hence, if f(x) = e*, ƒ'(x) = e*; and e* is a quantity which is equal to its own derived function. And similar results are true by reason of Article 31, is replaced by any function of x; hence we have when dy = ex {x2-2x+2} dx+e* {2x-2} dx, 35.] Differentiation of log, x. Suppose the base a of the last Article to be e, then since log, e = 1, And therefore if d.log, x = dx Hence the differential of the Napierian logarithm of any function of x is equal to the differential of the function divided by the function itself. 36.] The general result of Article 34 might also have been found from that of Article 32, as follows: whence we have an independent proof of the result of Art. 30. 37.] The differentiation of functions of a connected with one another by multiplication or division. y = f(x) × F(x) ×¢(x) × ................ If all the functions are positive, we have dy = = log y dy = = У ...... log f(x) + log r(x) + log p(x) + .............. ; d.f(x) d. F(x) d.p(x) + d{f(x) F(x) p (x) ... } = f(x) x (x) 4 (x) ... { d;} + $ (x) (d.f(x) d.F(x) + Or, if some of the functions are negative, + d.p(x) + + φ (α) ...}. [ƒ(x)]2 [F(x)]2 [4 (∞)]2 ................ ; log [f(x)]2+ log [F(x)]2 + log [p(x)]2 + = d.f(x) d. F(x) d.p(x) y f(x) + + F(x) f(x) ...... d{f(x) F(x) p(x)...} = d.f(x) F(x) (x)... +d.r(x) f(x)¢(x)..... +d.p(x) f(x) r(x)... +... (15) |