Second derivatives of r with x=r cosθ, y=r sinθ

Reading of the question. The scan was unclear about the right-hand side. I read it as printed: ∂²r/∂x² + ∂²r/∂y² = (1/r)[(∂r/∂x)² + (∂r/∂y)²], and this is a correct identity (both sides equal 1/r), so it is proved as printed.

Here x = r cos θ, y = r sin θ, so r² = x² + y², r = √(x² + y²).

First derivatives

Differentiating r² = x² + y²: 2r ∂r/∂x = 2x, so

∂r/∂x = x/r = cos θ
∂r/∂y = y/r = sin θ

Second derivatives

∂²r/∂x² = ∂/∂x (x/r) = 1/r − (x/r²)(∂r/∂x) = 1/r − x²/r³ = y²/r³
∂²r/∂y² = ∂/∂y (y/r) = 1/r − y²/r³ = x²/r³

Adding:

∂²r/∂x² + ∂²r/∂y² = (x² + y²)/r³ = r²/r³ = 1/r        ... (1)

Right-hand side

(∂r/∂x)² + (∂r/∂y)² = x²/r² + y²/r² = 1
(1/r)[(∂r/∂x)² + (∂r/∂y)²] = 1/r                       ... (2)

From (1) and (2):

∂²r/∂x² + ∂²r/∂y² = (1/r)[(∂r/∂x)² + (∂r/∂y)²]. ∎