finding mixed partials

Question

Let f(x,y) be a differentiable function with continuous partials and mixed partials. Let x(u,v)=ucos(v), y(u,v)=usin(v). show that fxx + fyy = (1/u)f_u+f_uu+(1/u²)f_vv.

Answer 1

This is an exercise in the chain rule and other differentiation rules, and in not making a mistake.

Therefore I will try to go slowly step by step.

I will use notation such as ∂f/∂x and f_x interchangeably.

1) Prepare the derivatives of x and y

x_u = cos(v)

x_v = -usin(v)

y_u = sin(v)

y_v = ucos(v)

x_uu = 0

x_vv = -ucos(v)

y_uu = 0

y_vv = -usin(v)

2) Calculate ∂f/∂u

f_u = f_x x_u + f_v y_u (by chain rule)

3) Calculate ∂²f/∂u² from f_u

f_uu = ∂/∂u(f_x) x_u + f_x ∂/∂u(x_u) + ∂/∂u(f_y) y_u + f_y ∂/∂u(y_u) (by rule of product)

= (f_xx x_u + f_xy y_u) x_u + f_x x_uu + (f_yx x_u + f_yy y_u) y_u + f_y y_uu (by chain rule)

= f_xx x_u² + f_xy y_u x_u + f_x x_uu + f_yx x_u y_u + f_yy y_u² + f_y y_uu (by distribution rule)

4) Calculate ∂f/∂v

f_v = f_x x_v + f_y y_v (by chain rule)

5) Calculate ∂²f/∂v² from f_v

f_vv = ∂/∂v(f_x) x_v + f_x ∂/∂v(x_v) + ∂/∂v(f_y) y_v + f_y ∂/∂v(y_v) (by rule of product)

= (f_xx x_v + f_xy y_v) x_v + f_x x_vv + (f_yx x_v + f_yy y_v) y_v + f_y y_v (by chain rule)

= f_xx x_v² + f_xy y_v x_v + f_x x_vv + f_yx x_v y_v + f_yy y_v² + f_y y_vv (by distribution rule)

6) Into the results of 2), 3), 5) substitute the results of 1)

f_u = f_x cos(v) + f_y sin(v)

f_uu = f_xx cos²(v) + f_xy sin(v)cos(v) + f_yx cos(v)sin(v) + f_yy sin²(v)

f_vv = f_xx u²sin²(v) - f_xy u²sin(v)cos(v) - f_x u cos(v) - f_yx u² sin(v)cos(v) + f_yy u²cos²(v) - f_y u sin(v)

5) Evaluate the given expression

I tried to align the term vertically to make the cancellations clearer.

(1/u) f_u + f_uu + (1/u²) f_vv =

(1/u) f_x cos(v) + (1/u) f_y sin(v) +

f_xx cos²(v) + f_xy sin(v)cos(v) + f_yx cos(v)sin(v) + f_yy sin²(v) +

f_xx sin²(v) - f_xy sin(v)cos(v) - (1/u) f_x cos(v) - f_yx sin(v)cos(v) + f_yy cos²(v) - (1/u) f_y sin(v) =

f_xx + f_yy