Second derivative of indetermined function

18 Views Asked by At

Suppose we have $g(x)$ and $g'(1)=g''(1)=1$

Now lets define $f(x,y)=xg(\frac{x}{y})$

And it is necessary to find $\frac{\delta^2 f}{\delta x^2}(1,1)+\frac{\delta^2 f}{\delta y^2}(1,1)$

So first I look for $\frac{\delta f}{\delta g}$, which is then:

$\frac{\delta f}{\delta g}=\frac{x(y-x\frac{dy}{dx})g'(x/y)}{y^2}+g(x/y)$

There should be no problem to get this derivated second time, if not for one thing which puzzles me: $\frac{dy}{dx}$

There is nothing said or implied about relation between $x$ and $y$. How to tackle $dy/dx$ in this case?