Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/124

Illustrative Example 2. Given $y=x^{2}e^{ax}$, find ${\tfrac {d^{n}y}{dx^{n}}}$ by Leibnitz's Formula.

Solution. Let	$u=x^{2}$,	and	$v=e^{ax}$;
then	${\frac {du}{dx}}=2x$	${\frac {dv}{dx}}=ae^{ax}$,
	${\frac {d^{2}u}{dx^{2}}}=2$,	${\frac {d^{2}v}{dx^{2}}}=a^{2}e^{ax}$,
	${\frac {d^{3}u}{dx^{3}}}=0$,	${\frac {d^{3}v}{dx^{3}}}=a^{3}e^{ax}$,
	.   .   .	.   .   .
	${\frac {d^{n}u}{dx^{n}}}=0$,	${\frac {d^{n}v}{dx^{n}}}=a^{n}e^{ax}$.

Substituting in (17), we get

${\frac {d^{n}y}{dx^{n}}}=x^{2}a^{n}e^{ax}+2na^{n-1}xe^{ax}+n(n-1)a^{n-2}e^{ax}=a^{n-2}e^{ax}[x^{2}a^{2}+2nax+n(n-1)]$.

78. Successive differentiation of implicit functions. To illustrate the process we shall find ${\frac {d^{2}y}{dx^{2}}}$ from the equation of the hyperbola

$b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}$.

Differentiating with respect to x, as in § 63,

$2b^{2}x-2a^{2}y{\frac {dy}{dx}}=0$,

or,

(A)	${\frac {dy}{dx}}={\frac {b^{2}x}{a^{2}y}}$.

Differentiating again, remembering that y is a function of x,

${\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}yb^{2}-b^{2}xa^{2}{\frac {dy}{dx}}}{a^{4}y^{2}}}$

Substituting for ${\frac {dy}{dx}}$ its value from (A),

${\frac {d^{2}y}{dx^{2}}}={\frac {a^{2}b^{2}y-a^{2}b^{2}x\left({\frac {b^{2}y}{a^{2}y}}\right)}{a^{4}y^{2}}}=-{\frac {b^{2}(b^{2}x^{2}-a^{2}y^{2})}{a^{4}y^{3}}}$.

But from the given equation, $b^{2}x^{2}-a^{2}y^{2}=a^{2}b^{2}$.

∴ ${\frac {d^{2}y}{dx^{2}}}=-{\frac {b^{4}}{a^{2}y^{3}}}$.