Let p : R^{2} → R^{2}, p(r, θ) = (rcosθ, rsinθ).

Then Dp(r, θ) = ( \begin{matrix}cosθ & −rsinθ \\ sinθ & rcosθ \end{matrix} )

Let f : R^{2} → R, f(x, y) = x^{2} + y^{2} + 3y (f ∘ p)(r, θ) = f(p(r, θ)) = f(rcosθ, rsinθ)

= r^{2}cos^{2}θ + r^{2}sin^{2}θ + 3rsinθ = r^{2} + 3rsinθ

Note we used the function p to convert x^{2} + y^{2} + 3y into polar coordinates. Use the chain rule to calculate D(f ∘ p)(r, θ): f(x, y) = x^{2} + y^{2} + 3y. Thus, Df(x, y) = ( 2x 2y + 3 ) D(f ∘ p)(r, θ) = (Df)(p(r, θ))Dp(r, θ)

= ( 2rcosθ 2rsinθ + 3 ) ( \begin{matrix}cosθ & −rsinθ \\ sinθ & rcosθ \end{matrix} ) =

( 2rcos^{2}θ + 2rsin^{2}θ + 3sinθ −2r^{2}sinθcosθ + 2r^{2}sinθcosθ + 3rcosθ)

= ( 2r + 3sinθ 3rcosθ)

D(f ∘ p)(r, θ) = (Df)(p(r, θ))Dp(r, θ)

= ( \frac{∂f}{∂x} \frac{∂f}{∂y} ) ( \begin{matrix}cosθ & −rsinθ \\ sinθ & rcosθ \end{matrix} )

= ( \frac{∂f}{∂x}cosθ + \frac{∂f}{∂y} sinθ −\frac{∂f}{∂x}rsinθ + \frac{∂f}{∂y} rcosθ )

\frac{∂}{∂r} = cosθ \frac{∂}{∂x} + sinθ \frac{∂}{∂y} \frac{∂}{∂θ} = −rsinθ \frac{∂}{∂x} + rcosθ \frac{∂}{∂y}

( \begin{matrix}cosθ & sinθ & \frac{∂}{∂r} \\ −rsinθ & rcosθ & \frac{∂}{∂θ} \end{matrix} ) → ( \begin{matrix}rcosθsinθ & rsin^{2}θ & rsinθ \frac{∂}{∂r} \\ −rcosθsinθ & rcos^{2}θ & cosθ \frac{∂}{∂θ} \end{matrix} )

→( \begin{matrix}rcosθsinθ & rsin^{2}θ & rsinθ \frac{∂}{∂r} \\ 0 & r & cosθ \frac{∂}{∂θ} + rsinθ \frac{∂}{∂r} \end{matrix} )

→( \begin{matrix}cosθ & sinθ & \frac{∂}{∂r} \\ 0 & 1 & \frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r} \end{matrix} )

→( \begin{matrix}cosθ & 0 & \frac{∂}{∂r} − sinθ[\frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r}] \\ 0 & 1 & \frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r} \end{matrix} )

→( \begin{matrix}cosθ & 0 & \frac{∂}{∂r} − sinθ \frac{cosθ}{r} \frac{∂}{∂θ} − sin^{2}θ \frac{∂}{∂r} \\ 0 & 1 & \frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r} \end{matrix} )

→( \begin{matrix}cosθ & 0 & −sinθ \frac{cosθ}{r} \frac{∂}{∂θ} + (1 − sin^{2}θ)\frac{∂}{∂r} \\ 0 & 1 & \frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r} \end{matrix} )→

( \begin{matrix}cosθ & 0 & −sinθ \frac{cosθ}{r} \frac{∂}{∂θ} + cos^{2}θ \frac{∂}{∂r} \\ 0 & 1 & \frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r} \end{matrix} )→ ( \begin{matrix}1 & 0 & \frac{−sinθ}{r} \frac{∂}{∂θ} + cosθ \frac{∂}{∂r} \\ 0 & 1 & \frac{cosθ}{r} \frac{∂}{∂θ} + sinθ \frac{∂}{∂r} \end{matrix} )

\frac{∂}{∂x} = cosθ \frac{∂}{∂r} − \frac{sinθ}{r} \frac{∂}{∂θ} \frac{∂}{∂y} = sinθ \frac{∂}{∂r} + \frac{cosθ}{r} \frac{∂}{∂θ}