Successive differentiation of implicit functions

To illustrate the process we shall find $\frac{d^2 y}{dx^2}$ from the equation of the hyperbola

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

Differentiating with respect to $x$, as in §5.35,

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

or,

$$\frac{dy}{dx} = \frac{b^2 x}{a^2 y}.$$ (7.2)

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 (7.2),

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

The given equation, $b^2x^2 - a^2y^2 = a^2b^2$, therefore gives,

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

Sage can be made to do a lot of this work for you (though the notation doesn't get any prettier):

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: x = var("x")
sage: y = function("y",x)
sage: a = var("a")
sage: b = var("b")
sage: F = x^2/a^2 - y^2/b^2 - 1
sage: F.diff(x)
2*x/a^2 - 2*y(x)*diff(y(x), x, 1)/b^2
sage: F.diff(x,2)
-2*y(x)*diff(y(x), x, 2)/b^2 - 2*diff(y(x), x, 1)^2/b^2 + 2/a^2
sage: solve(F.diff(x) == 0, diff(y(x), x, 1))
[diff(y(x), x, 1) == b^2*x/(a^2*y(x))]
sage: solve(F.diff(x,2) == 0, diff(y(x), x, 2))
[diff(y(x), x, 2) == (b^2 - a^2*diff(y(x), x, 1)^2)/(a^2*y(x))]

This basically says

$$y'=\frac{d y}{dx} = \frac{b^2x}{a^2 y},$$

and

$$y''=\frac{d^2 y}{dx^2} = - \frac{b^2-a^2(y')^2}{a^2 y}.$$

Now simply plug the first equation into the second, obtaining $y''=-b^2\frac{1-a^{-2}b^2x^2/y^2}{a^2y}$. Next, use the given equation in the form $a^{-2}b^2x^2/y^2-1=b^2/y^2$ to get the result above.