Higher Order Partial Derivatives — Question 3

Question 3

For $u(x,y)=\ln(x^2+y^2)$ on its natural domain:

Tasks

1. Find all second partial derivatives.

2. Show that $u_{xx}+u_{yy}=0$.

3. Explain why the result cannot be extended to the origin by substitution.

Question 3 – Solution

Strategy. Differentiate the rational first partials and preserve the punctured-plane domain.

Step 1: First partials $$u_x=\frac{2x}{x^2+y^2},\qquad u_y=\frac{2y}{x^2+y^2}.$$

Step 2: Second partials $$\boxed{u_{xx}=\frac{2(y^2-x^2)}{(x^2+y^2)^2}},\quad \boxed{u_{yy}=\frac{2(x^2-y^2)}{(x^2+y^2)^2}},$$ $$\boxed{u_{xy}=u_{yx}=-\frac{4xy}{(x^2+y^2)^2}}.$$

Step 3: Cancellation $$u_{xx}+u_{yy}=\frac{2(y^2-x^2+x^2-y^2)}{(x^2+y^2)^2}=\boxed 0.$$ The origin is excluded from $u$, and every denominator above vanishes there; substitution cannot create a derivative at a point outside the domain.