<s>[INST] Exercise 5.3 in Calculus Made Easy: are these answers equivalent? Exercise 5.3 in Calculus Made Easy, by Silvanus Thompson, is to find $\mathrm d\over \mathrm dx$ when the following relationship holds ($a$ and $b$ are both constants):
$$ay + bx = by - ax + (x + y)\sqrt{a^2 - b^2}$$
I tried differentiating both sides of this equation with respect to $x$ as follows:
$$\begin{align}
{\mathrm d\over \mathrm dx}(ay + bx) &= {\mathrm d\over \mathrm dx}\left(by - ax + (x + y)\sqrt{a^2 - b^2}\right)\\
a{\mathrm dy\over \mathrm dx} + b &= b{\mathrm dy\over \mathrm dx} - a + \left(1 + {dy\over dx}\right)\sqrt{a^2 - b^2}\\
\left(a - b - \sqrt{a^2 - b^2}\right){\mathrm dy\over \mathrm dx} &= \sqrt{a^2 - b^2} - a - b\\
{\mathrm dy\over \mathrm dx} &= {\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}
\end{align}$$
The author has considerably more finess and first does some algebraic manipulation to get the equation into better shape. First he squares both sides, then pushes some things around, then takes the square root of both sides:
$$\begin{align}
(a - b)^2y^2 + (a + b)^2x^2 + 2(a + b)(a - b)xy &= (x^2 + y^2 + 2xy)(a^2 - b^2)\\
\left[(a - b)^2 - \left(a^2 - b^2\right)\right]y^2 &= \left[\left(a^2 - b^2\right) - (a + b)^2\right]x^2\\
2b(b - a)y^2 &= -2b(b + a)x^2\\
y &= \sqrt{a + b \over a - b}x\\
{\mathrm dy\over \mathrm dx} &= \sqrt{a + b \over a - b}
\end{align}$$
I believe that both answers are correct, but if that is true, it should be possible to convert one to the other algebraically. For the life of me, I can't see how to do this. So my question is: are these two answers really equivalent?
 [/INST] $${\sqrt{a^2 - b^2} - a - b \over a - b - \sqrt{a^2 - b^2}}=\frac{\sqrt{a+b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}{\sqrt{a-b}\left(\sqrt{a-b}-\sqrt{a+b}\right)}=\frac{\sqrt{a+b}}{\sqrt{a-b}}$$
</s>