Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/123

If u and v are functions of x, we have, from V, §33,

${\frac {d}{dx}}(uv)={\frac {du}{dx}}v+u{\frac {dv}{dx}}$.

Differentiating again with respect to x,

${\frac {d^{2}}{dx^{2}}}(uv)={\frac {d^{2}u}{dx^{2}}}v+{\frac {du}{dx}}{\frac {dv}{dx}}+{\frac {du}{dx}}{\frac {dv}{dx}}+u{\frac {d^{2}v}{dx^{2}}}={\frac {d^{2}u}{dx^{2}}}v+2{\frac {du}{dx}}{\frac {dv}{dx}}+u{\frac {d^{2}v}{dx^{2}}}$

Similarly,

${\frac {d^{3}}{dx^{3}}}(uv)$	$={\frac {d^{3}u}{dx^{3}}}v+{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+2{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+2{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+u{\frac {d^{3}v}{dx^{3}}}$
	$={\frac {d^{3}u}{dx^{3}}}v+3{\frac {d^{2}u}{dx^{2}}}{\frac {dv}{dx}}+3{\frac {du}{dx}}{\frac {d^{2}v}{dx^{2}}}+u{\frac {d^{3}v}{dx^{3}}}$.

However far this process may be continued, it will be seen that the numerical coefficients follow the same law as those of the Binomial Theorem, and the indices of the derivatives correspond to the exponents of the Binomial Theorem.^[1] Reasoning them by mathematical induction from the mth to the (m + 1 )th derivative of the product, we can prove Leibnitz's Formula

${\frac {d^{n}}{dx^{n}}}(uv)={\frac {d^{n}u}{dx^{n}}}v+n{\frac {d^{n-1}u}{dx^{n-1}}}{\frac {dv}{dx}}+{\frac {n(n-1)}{2!}}{\frac {d^{n-2}u}{dx^{n-2}}}{\frac {d^{2}v}{dx^{2}}}+\cdots +$
$n{\frac {du}{dx}}{\frac {d^{n-1}v}{dx^{n-1}}}+u{\frac {d^{n}}{dx^{n}}}v$.

Illustrative Example 1. Given $y=e^{x}\ln x$, find ${\frac {d^{3}y}{dx^{3}}}$ by Leibnitz's Formula.

Solution. Let	$u=e^{x}$, and	$v=\log x$;
then	${\frac {du}{dx}}=e^{x}$,	${\frac {dv}{dx}}={\frac {1}{x}}$,
	${\frac {d^{2}u}{dx^{2}}}=e^{x}$,	${\frac {d^{2}v}{dx^{2}}}=-{\frac {1}{x^{2}}}$,
	${\frac {d^{3}u}{dx^{3}}}=e^{x}$	${\frac {d^{3}v}{dx^{3}}}={\frac {2}{x^{3}}}$.

Substituting in (17), we get

${\frac {d^{3}y}{dx^{3}}}=e^{x}\log x+{\frac {3e^{x}}{x}}-{\frac {3e^{x}}{x^{2}}}=e^{x}\left(\log x+{\frac {3}{x}}-{\frac {3}{x^{2}}}+{\frac {2}{x^{3}}}\right)$.

1. To make this correspondence complete, u and v are considered as ${\frac {d^{0}u}{dx^{0}}}$ and ${\frac {d^{0}v}{ddx^{0}}}$.