I've the following equation: $b^2x^2 + a^2y^2 = a^2b^2$, the first implicit derivative is:
$\dfrac{dy}{dx} = \dfrac{-b^2x}{a^2y}$
I do not undertand how to find the second derivative of this equation, could someone guide me ?
Ps: I guess the second derivative is: $\dfrac{d}{dx}(\dfrac{-b^2x}{a^2y}) = \dfrac{b^2*(x-y)}{a^2y^2}$, but isn't right.
Is it the right asnwer: $\dfrac{(-a^2b^2y^2) - (b^4x^2)}{(a^4y^3)}$ ?
On
$$b^2x^2 + a^2y^2 = a^2b^2 \Rightarrow \frac{b^2}{a^2} = \frac{b^2 - y^2}{x^2}$$
Then:
$$\dfrac{dy}{dx} = \dfrac{-b^2x}{a^2y} = -\frac{(b^2-y^2)x}{x^2y} = \frac{y^2-b^2}{xy}$$
At this point, you can separate variables:
$$dy \frac{y}{y^2-b^2} = dx \frac{1}{x}$$
Integrating...
$$\int_{y_0}^y \frac{t}{t^2-b^2} dt = \int_{x_0}^x\frac{1}{s}ds$$ $$\frac{1}{2}\int_{y_0}^y \frac{2t}{t^2-b^2} dt = \log(x) - \log(x_0)$$ $$\frac{1}{2}[\log(y^2-b^2) - \log(y_0^2 -b^2)] = \log(x) - \log(x_0)$$ $$\frac{1}{2}\log(y^2-b^2) = \log\left(\frac{x}{x_0}\right) + \frac{1}{2}\log(y_0^2 -b^2)$$ $$\log(\sqrt{y^2-b^2}) = \log\left(\frac{x\sqrt{y_0^2-b^2}}{x_0}\right) $$ $$\sqrt{y^2-b^2} = \left(\frac{x\sqrt{y_0^2-b^2}}{x_0}\right) $$ $$y^2 = \left(\frac{x\sqrt{y_0^2-b^2}}{x_0}\right) + b^2$$ $$y(x) = \sqrt{\left(\frac{x\sqrt{y_0^2-b^2}}{x_0}\right) + b^2}$$
where
$$y(x_0) = y_0$$
It's best to differentiate implicitly twice. If $b^2x^2 + a^2y^2 = a^2b^2$ then $$2b^2x+2a^2yy' = 0$$ $$2b^2 + 2a^2(y')^2 + 2a^2yy'' = 0$$ As you have done: solve $2b^2x+2a^2yy' = 0$ for $y'$.
Substitute your expression for $y'$ into $2b^2 + 2a^2(y')^2 + 2a^2yy'' = 0$ and solve for $y''$.
Note: Your guess at the second derivative doesn't work because $y$ is a function of $x$. If you have $$y' = -\frac{b^2}{a^2}\left(\frac{x}{y}\right)$$ then differentiating with respect to $x$ gives (by applying the quotient rule) $$y'' = -\frac{b^2}{a^2}\left(\frac{y-xy'}{y^2}\right)$$