Partial differentiation chain rule, differential operator?

Question

We are given the function \begin{equation} V(x,y)= f(s)+g(t) \end{equation} with s=x+y and t=x+0.5y.

How can I calculate $V_{xx}$ and $V_{yy}$? I have calculated $V_{x}$ and $V_{y}$ but I do not know how to apply the differential operator to the derivatives. \begin{equation} \frac{\partial{V}}{\partial{x}}=\frac{\partial{V}}{\partial{s}} \frac{\partial{s}}{\partial{x}}+\frac{\partial{V}}{\partial{t}} \frac{\partial{t}}{\partial{x}}=\frac{\partial{V}}{\partial{s}} +\frac{\partial{V}}{\partial{t}} \end{equation} Hence, \begin{equation} V_{xx}=\frac{\partial{}V_{x}}{\partial{x}}=\frac{\partial{}}{\partial{x}}(\frac{\partial{V}}{\partial{s}} +\frac{\partial{V}}{\partial{t}})=(\frac{\partial{}}{\partial{s}} +\frac{\partial{}}{\partial{t}})(\frac{\partial{V}}{\partial{s}} +\frac{\partial{V}}{\partial{t}}) \end{equation} where I have plugged in the differential operator. Then, however I would also get mixed derivatives. \begin{equation} (\frac{\partial{}}{\partial{s}} +\frac{\partial{}}{\partial{t}})(\frac{\partial{V}}{\partial{s}} +\frac{\partial{V}}{\partial{t}})=V_{ss}+V_{st}+V_{ts}+V_{tt} \end{equation}

Assuming this method is correct, how do I get rid of the mixed derivatives?

Answer 1

We have $V(x,y)=f(x+y)+g(x+0.5y)$ so that $$V_y(x,y)=f'(x+y)+0.5g'(x+0.5y)$$ and $$V_{yy}(x,y)=f''(x+y)+0.25g''(x+0.5y)$$ Finding $V_{xx}$ is even easier.

Answer 2

$$V(x,y)= f(s)+g(t)$$ Then $$V_x=\frac{\partial s}{\partial x}f'(s)+\frac{\partial t}{\partial x}g'(t)$$ And so on. This is how you get factors of $1/2$ each time you differentiate $g$ by $y$.

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If, say $f$ had been a more complicated function, $f=f(s,t)$, then you'd have $$V_x=\frac{\partial s}{\partial x}\frac{\partial f}{\partial s}+\frac{\partial t}{\partial x}\frac{\partial f}{\partial t}+\frac{\partial t}{\partial x}g'(t)$$