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



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
 * $$\frac{du}{dx} = 2x$$
 * $$\frac{dv}{dx} = ae^{ax}$$,
 * $$\frac{d^2 u}{dx^2} = 2$$,
 * $$\frac{d^2 v}{dx^2} = a^2 e^{ax}$$,
 * $$\frac{d^3 u}{dx^3} = 0$$,
 * $$\frac{d^3 v}{dx^3} = a^3 e^{ax}$$,
 * style="text-align: right;"| .  ..
 * $$\frac{d^n u}{dx^n} = 0$$,
 * $$\frac{d^n v}{dx^n} = a^n e^{ax}$$.
 * }
 * $$\frac{d^3 v}{dx^3} = a^3 e^{ax}$$,
 * style="text-align: right;"| .  ..
 * $$\frac{d^n u}{dx^n} = 0$$,
 * $$\frac{d^n v}{dx^n} = a^n e^{ax}$$.
 * }
 * $$\frac{d^n u}{dx^n} = 0$$,
 * $$\frac{d^n v}{dx^n} = a^n e^{ax}$$.
 * }
 * $$\frac{d^n v}{dx^n} = a^n e^{ax}$$.
 * }


 * Substituting in (17), we get

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,

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


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

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, $$b^2 x^2 - a^2 y^2 = a^2 b^2$$.


 * &there4; $$\frac{d^2 y}{dx^2} = - \frac{b^4}{a^2 y^3}$$.