There are lots of examples of and questions about partial derivatives for specific functions. However, I'm looking at a generic form:

Let's say $w(x,y)=f(mx-ny)+g(mx+ny)$

Taking the derivative with respect to $y$, I would think:

$\frac{\partial w}{\partial y}=\frac{\partial}{\partial y}f(mx-ny)(-n)+\frac{\partial}{\partial y}g(mx-ny)(n)$

I'm not sure whether I need to use the chain rule because of the two variables.

I'm trying to find $\frac{\partial^2w}{\partial y^2}$


Your thinking is correct. Perhaps this will give you confidence about it.

One way to think of computing this partial derivative is the following:

You have a function $\alpha(x, y) = mx- ny$ and $\beta(x, y) = mx + ny$, which take points in $\mathbb{R}^2$ to $\mathbb{R}$ (i.e. we are mapping points in the plane to to real numbers). You implicitly define the functions $f$ and $g$ from $\mathbb{R}$ to $\mathbb{R}$ (i.e. we are mapping real numbers to real numbers).

Thus, we can compose $\alpha$ and $\beta$ with $f$ and $g$ to get $w$.

Set $w(x, y) = f(\alpha(x, y)) + g(\beta(x, y)) = f(mx - ny) + g(mx + ny)$

Hence, the application of the chain rule is clearer:

$\frac{\partial w}{\partial y}(x,y) = f'(\alpha(x, y))\frac{\partial \alpha}{\partial y}(x, y) + g'(\beta(x,y))\frac{\partial \beta}{\partial y}(x,y) = f'(mx - ny)(-n) + g'(mx + ny)(n)$

Computing $\frac{\partial^2 w}{\partial y^2}$ is very similar -- just be sure to apply the product rule.


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  • $\begingroup$ Product rule would yield: f"(mx-ny)(n^2)+g"(mx+ny)(n^2) or n^2(f"(mx-ny)+g"(mx+ny))? $\endgroup$ – johnnd Jul 25 '17 at 18:16
  • $\begingroup$ The two answers you just gave are the same, but correct. $\endgroup$ – Hunter Vallejos Jul 25 '17 at 18:31

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