the function:
$$\mathrm{ln}\sqrt{x^2+y^2} = \mathrm{arctg}\frac{y}{x}$$
After finding first derivative I have (took $y$ as a complicated func):
$$y' = \frac{x^2}{x^2-yx-y^2}$$
but it does not seem to be correct answer, started to learn implicit funcs just few days ago, can someone give me the proper direction with its derivatives?
From $$\ln (x^2+y^2) = 2\arctan\frac{y}{x}$$ we get on differentiating, $$\frac{x + yy'}{x^2+y^2} = \frac{1}{1+y^2/x^2} \frac{xy' - y}{x^2}$$ Simplifying, we get $$x+yy' = xy' - y$$ ad hence $$y' = \frac{x+y}{x-y}$$ Differentiate again: $$y'' = \frac{(1+y')(x-y) - (1-y')(x+y)}{(x-y)^2}$$ Substitute for $y'$ and simplify to obtain $$y'' = \frac{2(x^2+y^2)}{(x-y)^3}$$