Find polar function, knowing the 'Cartesian' derivative

I know the Cartesian derivative and would like to find the formula for the polar function. I am not sure whether my thinking is correct, however.

$$\frac{dy}{dx}=\tan\left(\theta+\frac{5}{4}\pi\right)$$ ...this much I know.

Now my thinking would be that since:
$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

I'd have:

$$\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\tan\left(\theta+\frac{5}{4}\pi\right)=\frac{c\sin\left(\theta+\frac{5}{4}\pi\right)}{c\cos\left(\theta+\frac{5}{4}\pi\right)}$$

so:

$$\frac{dy}{d\theta}=c\sin\left(\theta+\frac{5}{4}\pi\right), \space \frac{dx}{d\theta}=c\cos\left(\theta+\frac{5}{4}\pi\right)$$

Now I doubt this is correct. I mean, there is plenty of other possible pairs of functions. But then, if this isn't correct, is there any other way to find the polar function formula by just having the tangent in Cartesian coordinates?

I kind of feel like I'm approaching this problem from the wrong side, but also it feels like it should be possible to find the formula for the polar function, knowing the tangent, up to maybe some multiplication/addition (the same way as in Cartesian). Obviously, I have close to zero experience with polar functions, so any kind of advice will be greatly appreciated.

Answer

Well! $c$ itself can be a function of $\theta$ so this approach is incorrect. But the following way leads to something better i think:$$\dfrac{dy}{dx}=\dfrac{1+\tan\theta}{1-\tan\theta}=\dfrac{1+\dfrac{y}{x}}{1-\dfrac{y}{x}}$$Let $y=xu$ therefore $$\dfrac{dy}{dx}=u+xu'=\dfrac{1+u}{1-u}\to\\xu'=\dfrac{1+u^2}{1-u}\to\dfrac{dx}{x}=\dfrac{1-u}{1+u^2}du\to\\\ln x+C_1=\tan^{-1}u-\dfrac{1}{2}\ln({1+u^2})\to\\C_2x=\dfrac{e^{\tan^{-1}\frac{y}{x}}}{\sqrt{1+\dfrac{y^2}{x^2}}}$$or$$C_2r\cos\theta=\dfrac{e^{\theta}}{\sqrt{1+\tan^2\theta}}$$which yields to $$r=Ce^\theta$$