<s>[INST] Prove derivative of an implicit function The question:
$\sqrt{1+x^2} + \sqrt{1+y^2} = a(x-y)$ prove that $\dfrac{dy}{dx} = \sqrt{\dfrac{1+y^2}{1+x^2}}$
I tried doing it the normal way and got
$\dfrac{dy}{dx} = \dfrac{(a\sqrt{1+x^2}-x)(\sqrt{1+y^2})}{(a\sqrt{1+y^2} + y)(\sqrt{1+x^2})}$
I also tried doing a trigonometry sub to get a relation of $a$ into a independent constant
$x=\tan(\alpha)$
$y=\tan(\beta)$
And got
$$a = \dfrac{\cos(\alpha) + \cos(\beta)}{\sin(\alpha - \beta)}$$ I don't know if this one can be further manipulated.
Is there something I missed? or is there another way of solving it?
 [/INST] Note that $$a=\frac{\sqrt{1+x^{2}}+\sqrt{1+y^{2}}}{x-y}.$$
so that
$$
\begin{align*}
\frac{a\sqrt{1+x^{2}}-x}{a\sqrt{1+y^2}+y} &=\frac{\frac{(1+x^{2})+\sqrt{1+x^{2}}\sqrt{1+y^{2}}-x^2+xy}{x-y}}{\frac{(1+y^{2})+\sqrt{1+x^{2}}\sqrt{1+y^2}+xy-y^{2}}{x-y}}\\
&=\frac{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}\\
&=1.
\end{align*}$$
Therefore from your own calculations
$$\frac{dy}{dx}=\frac{(a\sqrt{1+x^{2}}-x)(\sqrt{1+y^{2}})}{(a\sqrt{1+y^{2}}+y)(\sqrt{1+x^{2}})}=\frac{\sqrt{1+y^{2}}}{\sqrt{1+x^{2}}}.$$
</s>