The question is $\Rightarrow $ ${\sqrt {1-x^2}}+{\sqrt {1-y^2}} =a (x-y)$. Show that $\frac{dy}{dx}=\left(\frac{\sqrt {1-y^2}}{\sqrt{1-x^2}}\right)$
Below I've mentioned how i approached it.
$${\sqrt {1-x^2}}+{\sqrt {1-y^2}} =a (x-y) \Rightarrow (1)$$ $$let; 1-x^2=u^2, 1-y^2=v^2$$ $$u+v=a ({\sqrt {1-u^2}}-{\sqrt{1-v^2}})$$ differentiating with respect to v
$$\frac{du}{dv}=\left(\frac{\sqrt {1-u^2}}{\sqrt{1-v^2}}\right)\left(\frac {av-{\sqrt {1-v^2}}} {au+{\sqrt{1-u^2}}}\right)$$ $${when;u\to y},{v\to x} \Rightarrow \alpha$$ $$\frac{dy}{dx}=\left(\frac{\sqrt {1-y^2}}{\sqrt{1-x^2}}\right)\left(\frac {ax-{\sqrt {1-x^2}}} {ay+{\sqrt{1-y^2}}}\right)$$
From (1) the things in second brackets cancel of, $$Finally,\\ \frac{dy}{dx}=\left(\frac{\sqrt {1-y^2}}{\sqrt{1-x^2}}\right)$$
The only reason for using step $α$ is because when done the things in the brackets cancel off.
Since that is wrong if there is any other method (than the one where you substitute sinα for x and sinβ for y and so on...) let me know.
Also can anyone explain to me when and where i can use step α and why i cant use it wherever i want to.
The method where you substitute sinα & sinβ for x and y.Maybe It'll be helpfull for someone else.
$${\sqrt {1-x^2}}+{\sqrt {1-y^2}} =a (x-y) \\ let;x=\sin α, y=\sin β\\then; α=\arcsin x , β=\arcsin y$$ Substitute for $x$ & $y$ $${\sqrt {1- (\sin α)^2}}+{\sqrt {1- (\sinβ)^2}} =a (\sinα-\sinβ) \\ \cosα+\cosβ=a (\sinα -\sinβ) \\ 2\cos\left(\frac {α+β}{2}\right)\cos\left(\frac {α-β}{2}\right)=a \biggl(2\cos\left(\frac {α+β}{2}\right)\sin\left(\frac {α-β}{2}\right)\biggr) \\ \frac {1}{a} = \tan\left(\frac {α-β}{2}\right) \\ 2\arctan \left(\frac{1}{a}\right)= α-β$$ Substitute for α & β
$$2\arctan \left(\frac{1}{a}\right)=\arcsin x - \arcsin y $$
By Differentiating with respect to x $$0= \left(\frac {1}{\sqrt {1-x^2}}\right)-\frac {dy}{dx}\left(\frac {1}{\sqrt {1-y^2}}\right) \\ \frac {dy}{dx}=\left(\frac {\sqrt {1-y^2}}{\sqrt {1-x^2}}\right) $$ $\Bbb Note: $ $2\arctan \left(\frac{1}{a}\right)$ is a constant so when differentiated with respect to x it equals to zero
$ \frac {d}{dx} 2\arctan \left(\frac{1}{a}\right)=0$