Second derivative of $\dfrac {x-y}{x+y}$

Question

I‘m trying to find the second derivative of

$f:\left( 0,\infty \right) \times \left( 0,\infty \right) \rightarrow \mathbb{R}, \left( x,y\right) \mapsto \dfrac {x-y}{x+y}$.

I already managed to calculate the partial derivatives

$\partial _{x}f\left( x,y\right) =-\partial _{y}f\left( x,y\right) =\dfrac {2y}{\left( x+y\right) ^{2}}$.

The partial derivatives are continuous on $\left( 0,\infty \right) \times \left( 0,\infty \right)$, so $f$ is totally differentiable. But what is the total derivative of $f$? And how do i go from there?

Answer 1

You didn't calculate $\partial_y f(x,y)$ correctly, it should be $-2x/(x+y)^2$.

Total derivative is linear operator $\mathbb R^2\to \mathbb R$ which is represented by the matrix

\begin{bmatrix} \partial_x f(x,y)& \partial_y f(x,y) \end{bmatrix}

in the standard basis.

The Hessian, then, is given by

\begin{bmatrix} \partial_x\partial_x f(x,y)& \partial_x\partial_y f(x,y)\\ \partial_y\partial_x f(x,y)& \partial_y\partial_y f(x,y) \end{bmatrix}