Linearising an implicit relation in respect to one of the variables

Question

I have a function of this form $$ x + y = \frac{x + ay}{b^2 \sqrt{(x + ay)^2+c^2}}, $$ and I would like to find a linear approximation for $x$ in the form $$ x = y_0 + dy + \mathcal{O}(x^2), $$ what is the way to do it?

Answer 1

Well it goes like this $$ x_0 + \delta x + \delta y = \frac{(x_0 + \delta x+a \delta y)}{b^2\sqrt{(x_0 + \delta x+a \delta y)^2 + c^2}}\\ (x_0 + \delta x + \delta y){b^2\sqrt{(x_0 + \delta x+a \delta y)^2 + c^2}} = {(x_0 + \delta x+a \delta y)}\\ (x_0 + \delta x + \delta y){b^2\sqrt{x_0^2 + 2x_0 \delta x + 2 x_0 a \delta y + c^2}} = {(x_0 + \delta x+a \delta y)}\\ (x_0 + \delta x + \delta y){b^2\sqrt{(x_0^2 + c^2) + \frac{(x_0^2 + c^2)2x_0 (\delta x + a \delta y)}{(x_0^2 + c^2)}}} = {(x_0 + \delta x+a \delta y)}\\ (x_0 + \delta x + \delta y){b^2\sqrt{(x_0^2 + c^2) + \frac{(x_0^2 + c^2)2x_0 (\delta x + a \delta y)}{(x_0^2 + c^2)}}} = {(x_0 + \delta x+a \delta y)}\\ (x_0 + \delta x + \delta y){b^2\sqrt{(x_0^2 + c^2)}\sqrt{1 + \frac{2x_0 (\delta x + a \delta y)}{(x_0^2 + c^2)}}} = {(x_0 + \delta x+a \delta y)} $$ and as we have $$x_0 = \frac{x_0}{b^2 \sqrt{x_0^2 + c^2}} $$ we get to $$(x_0 + \delta x + \delta y){\sqrt{1 + {2x_0 {b^4} (\delta x + a \delta y)}}} = {(x_0 + \delta x+a \delta y)}\\ (x_0^2 + 2 x_0 \delta x + 2 x_0 \delta y){(1 + {2x_0 {b^4} (\delta x + a \delta y)})}= {x_0^2 + 2 x_0 \delta x+ 2 x_0 a \delta y} \\ {(x_0^2 + 2 x_0 \delta x + 2 x_0 \delta y) + {2x_0 (\delta x + a \delta y)}{b^4}(x_0^2 + 2 x_0 \delta x + 2 x_0 \delta y)}= {x_0^2 + 2 x_0 \delta x+ 2 x_0 a \delta y} \\ { 2 x_0 \delta y + {2x_0 (\delta x + a \delta y)}{b^4}x_0^2}= { 2 x_0 a \delta y} \\ {\delta y + {{b^4}x_0^2 \delta x + {b^4}x_0^2 a \delta y}}= { a \delta y} \\ { {{b^4}x_0^2 \delta x }}= ( a - 1 - {b^4}x_0^2 a) \delta y\\ \delta x = \frac{( a - 1 - {b^4}x_0^2 a)}{{{b^4}x_0^2}} \delta y $$