# Calculus/Polar Differentiation

### Differential calculusEdit

We have the following formulae:

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

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

${\displaystyle x=r(\theta )\cos(\theta )}$
${\displaystyle y=r(\theta )\sin(\theta )}$

Differentiating both equations with respect to ${\displaystyle \theta }$  yields

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

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

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