Determine the first and second orders derivatives of $y = f(x)$ determined implicity by

$$ \ln\sqrt{x^2 + y^2} = \alpha~\arctan\frac{y}{x} $$

Now, notice that in polar coordinates, the expression for this equation becomes simply

$$ \ln (r) = \alpha \theta $$

Which is much easier to deal with, especially when calculating a second order derivative, since higher order derivatives tend to become "hairy".

So I want to use polar coordinates to solve this problem but now I'm in doubt: is $\theta$ an implicit function of $r$ or is it the other way?

  • $\begingroup$ I don't see why you need to use polar co-ordinates at all - Just differentiate both sides w.r.t. $x$ and use the chain rule. $\endgroup$ Commented Apr 25, 2019 at 20:34
  • $\begingroup$ I don't need to. In fact I calculated the first order derivative, however the expression is big and calculating the second order derivative would take forever in an exam. $\endgroup$
    – Sigma
    Commented Apr 25, 2019 at 20:36
  • $\begingroup$ Perhaps, but if you're asked to express it in the form $\frac{\partial y}{\partial x} = f(x)$ and $\frac{\partial^2 y}{\partial x^2} = f'(x)$, then you'll have to use the chain rule ultimately anyway, and I doubt that your expression will be much simpler. $\endgroup$ Commented Apr 25, 2019 at 20:45
  • $\begingroup$ I do not get your question, it's a multi-variable function so both $r$ and $\theta$ are variables. $\endgroup$
    – Vasili
    Commented Apr 25, 2019 at 21:17

1 Answer 1


$$ \ln\sqrt{x^2 + y^2} = \alpha~\arctan\frac{y}{x} $$

The implicit calculus is not so difficult than appears at first sight. We differentiate both terms :

$$\frac{x\:dx+y\:dy}{x^2+y^2}=\alpha\frac{x\:dx-y\:dy}{x^2+y^2}$$ After simplification : $$\boxed{y'=\frac{dy}{dx}=\frac{x+\alpha y}{\alpha x-y}}$$ We differentiate again :

$d(y')=\frac{\partial}{\partial x}\left(\frac{x+\alpha y}{\alpha x-y}\right)dx+\frac{\partial}{\partial y}\left(\frac{x+\alpha y}{\alpha x-y}\right)dy$

$d(y')=-\frac{(1+\alpha^2)y}{(\alpha x-y)^2}dx+\frac{(1+\alpha^2)x}{(\alpha x-y)^2}dy$

$d(y')=-\frac{(1+\alpha^2)y}{(\alpha x-y)^2}dx+\frac{(1+\alpha^2)x}{(\alpha x-y)^2}\frac{x+\alpha y}{\alpha x-y}dx$

$d(y')=(1+\alpha^2)\frac{ -y(\alpha x-y)+x(x+\alpha y)}{(\alpha x-y)^3}dx$ $$\boxed{y''=\frac{d^2y}{dx^2}=\frac{d(y')}{dx}=(1+\alpha^2)\frac{x^2+y^2}{(\alpha x-y)^3}}$$


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