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 x 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{(x)}. Since if y = f(x), dy=f'(x) dx; therefore, if x is replaced by p(x), da must be replaced by the differential of (x), that is, by p'(x) dx; and therefore if y = ƒ{p(x)}, dy = ƒ'{$(x)} p′(x) dx. And so again, if the x involved in (x) is a function of x, say (x), (x) will be replaced by {(x)}, and '(x) dx by ☀'{4(x)} \'(x) dx; and a similar notation will adapt itself to a function of a function...... (to n functions) of x. Thus if y = ƒ{d[4(x)]}, dy = ƒ'{$[¥(x)]} 4′[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)1 (x+3)2 (x2-12x-6) (x2-2)3 dx. Let a-1=z, so that ar and z are simultaneously infinitesimal; .. aax = 2+1, ▲x loge a = loge (1 + 2), and by Cor. I, Lemma I, log, (1 + z) = z, when z is infinitesimal; and therefore adx-1 must be replaced by dx log, a ; 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 a* = 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 r 35.] Differentiation of log, x. Suppose the base a of the last Article to be e, then since 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 x connected with one another by multiplication or division. y = f(x) × F(x) × $(x) × If all the functions are positive, we have log y = ...... log f(x) + log F(x) + log (x) + ...... ; dy d.f(x) d. F(x) d.p(x) = + y f(x) F(x) d{f(x) F(x) p(x) ..... } + φ (*) + ...... Or, if some of the functions are negative, y2 = [f(x)]3 [r(x)]2 [$ (∞)]2 ............·· ; log [f(x)]2+ log [F(x)]2 + log [p(x)]2 + log (y2) = · · dy = f(x) x (x) p (x) ... { d.f(x) + d.r(x) + d.p(x) + Therefore, in either case, f(x) F(x) φ (α) ...... ...}. d{f(x) F(x) p(x)...} = d.f(x) F(x) (x)... +d.F(x) f(x)+(x)....... |