# Calculus/Polar Differentiation

### Differential calculus

We have the following formulae:

$r{\frac {\partial }{\partial r}}=x{\frac {\partial }{\partial x}}+y{\frac {\partial }{\partial y}}$
${\frac {\partial }{\partial \theta }}=-y{\frac {\partial }{\partial x}}+x{\frac {\partial }{\partial y}}$

To find the Cartesian slope of the tangent line to a polar curve $r(\theta )$  at any given point, the curve is first expressed as a system of parametric equations.

$x=r(\theta )\cos(\theta )$
$y=r(\theta )\sin(\theta )$

Differentiating both equations with respect to $\theta$  yields

${\frac {\partial x}{\partial \theta }}=r'(\theta )\cos(\theta )-r(\theta )\sin(\theta )$
${\frac {\partial y}{\partial \theta }}=r'(\theta )\sin(\theta )+r(\theta )\cos(\theta )$

Dividing the second equation by the first yields the Cartesian slope of the tangent line to the curve at the point $(r,r(\theta ))$  :

${\frac {dy}{dx}}={\frac {r'(\theta )\sin(\theta )+r(\theta )\cos(\theta )}{r'(\theta )\cos(\theta )-r(\theta )\sin(\theta )}}$