$x=\cos \theta - r \sin \theta, \ y = \sin \theta + r \cos \theta $ Squaring and adding, $x^2+y^2 = (\cos \theta - r \sin \theta)^2 + (\sin \theta + r \cos \theta)^2$ $ x^2+y^2 = \cos^2 \theta - 2r \sin \theta \cos \theta + r^2 \sin^2 \theta + \sin^2 \theta + 2r \sin \theta \cos \theta + r^2 \cos^2 $ $x^2+y^2= (\cos^2 \theta + \sin^2 \theta)+r^2 (\cos^2 \theta + \sin^2 \theta) = 1+r^2$ $Implicitly \ differentiating \ x^2+y^2=1+r^2, \ w.r.t\ x,$ $2x+0=0 + \dfrac {2rdr}{dx} \ \therefore \dfrac {dr}{dx}= \dfrac {x}{r}$