Deriving the addition formula for the lemniscate functions from a total differential equation

The following proof is basically the same as the previous one, but some people may prefer this.

Let $u + v = c$, where $c$ is a constant. It suffices to prove that

$$s(c) = \frac{s(u)c(v) + s(v)c(u)}{1 - s(u)s(v)c(u)c(v)}$$

Since $v = c - u$, the right hand side of this equation will be

$$f(u) = \frac{s(u)c(c - u) + s(c - u)c(u)}{1 - s(u)s(c - u)c(u)c(c - u)}$$

Suppose $f'(u) = 0$. Then $f(u)$ is a constant. Hence $f(u) = f(0)$. On the other hand, since $s(0) = 0$ and $c(0) = s(\omega) = 1$, $f(0) = s(c)$. Hence $f(u) = s(c)$ and we are done.

So it suffices to prove that $f'(u) = 0$. This is only a tedious routine.

We follow the method of my answer to this question.

Let $u = \int_{0}^{x}\frac{dx}{\sqrt{1 - x^4}}$. Then $x = s(u)$.

Let $v = \int_{0}^{y}\frac{dy}{\sqrt{1 - y^4}}$. Then $y = s(v)$.

Let $c$ be a constant. $u + v = c$ is a solution of the equation:

$$\frac{dx}{\sqrt{1 - x^4}} + \frac{dy}{\sqrt{1 - y^4}} = 0$$

It suffices to prove that $s(c) = \frac{x\sqrt{1 - y^4} + y\sqrt{1 - x^4}}{1 + x^2y^2}$. Since $v = c - u$, the right hand side is a function of $u$. We write this function by $\phi(u)$. Namely, $$\phi(u) = \frac{x\sqrt{1 - y^4} + y\sqrt{1 - x^4}}{1 + x^2y^2}$$

Let us compute $\frac{d\phi}{du}$.

$\frac{dx}{du} = 1/\frac{du}{dx} = \sqrt{1 - x^4}$

$\frac{dy}{du} = -\frac{dy}{dv} = -1/\frac{dv}{dy} = -\sqrt{1 - y^4}$

$\frac{d^2x}{du^2}= \frac{d\sqrt{1 - x^4}}{du}\cdot\frac{dx}{du} = \frac{-2x^3}{\sqrt{1 - x^4}} \sqrt{1 - x^4} = -2x^3$

$\frac{d^2y}{du^2}= \frac{d^2y}{dv^2} = -2y^3$

Hence $\frac{d\phi}{du} = \frac{d}{du}(\frac{-x\frac{dy}{du} + y\frac{dx}{du}}{1 + x^2y^2})$

Let $f = -x\frac{dy}{du} + y\frac{dx}{du}$

Let $g = 1 + x^2y^2$

Then $\frac{d}{du}(\frac{f}{g}) = \frac{\frac{df}{du}}{g} - \frac{f\frac{dg}{du}}{g^2}$

$\frac{df}{du} = -\frac{dx}{du}\frac{dy}{du} - x\frac{d^2y}{du^2} + \frac{dy}{du}\frac{dx}{du} + y\frac{d^2x}{du^2} = y\frac{d^2x}{du^2} - x\frac{d^2y}{du^2} = 2(xy^3 - yx^3)$

$\frac{dg}{du} = 2xy^2\frac{dx}{du} + 2x^2y\frac{dy}{du} = 2xy(y\frac{dx}{du} + x\frac{dy}{du})$

$f\frac{dg}{du} = 2xy(y^2(\frac{dx}{du})^2 - x^2(\frac{dy}{du})^2) = 2xy(y^2(1 - x^4) - x^2(1 - y^4))$

Hence $\frac{d}{du}(\frac{f}{g}) = \frac{2(xy^3 - yx^3)}{1 + x^2y^2} - \frac{2xy(y^2(1 - x^4) - x^2(1 - y^4)}{(1 + x^2y^2)^2} = \frac{2(1 + x^2y^2)(xy^3 - yx^3) - 2xy(y^2(1 - x^4) - x^2(1 - y^4)}{(1 + x^2y^2)^2} = 0$

Hence $\frac{d\phi}{du} = 0$

Hence $\phi(u)$ is constant. Hence $\phi(u) = \phi(0) = y = s(v) = s(c)$ as desired