Given the function $z=f(\sqrt{x^2+y^2})$, calculate it's second (total) derivative.

I've already calculated the first order derivative and it is equal to:

$$dz=f' \frac{xdx+ydy}{\sqrt{x^2+y^2}}$$

I'm having trouble proceeding further. In the book the end result is:

$$d^2u=d(f') \frac{xdx+ydy}{\sqrt{x^2+y^2}} + f'd\left(\frac{xdx+ydy}{\sqrt{x^2+y^2}}\right)=f'' \frac{(xdx+ydy)^2}{x^2+y^2}+f' \frac{(ydx-xdy)^2}{\sqrt{(x^2+y^2)^3}}$$

Which looks a lot like single variable product rule... I don't think I can use the chain rule again simply because this is not a composition of functions anymore, but it's more like a product of functions. Also $f$ is unknown function, in fact the only thing I know about it is that it's a single variable function. So even if I try to calculate partial derivatives here I don't even know all the variables ($x$ and $y$ are the main ones, but also $f'$ is a function of one variable). I guess in the end $f$ is calculated through $\sqrt{x^2+y^2}$ so it's function dependent on $x$ and $y$ and taking the partial derivatives with respect to $x$ and $y$ is the correct way.

Forgive me if I'm missing something very obvious. I'm quite new to this subject and more detail will be appreciated.

Also how do we treat the $dx$ and $dy$ part when calculating higher order derivatives? They represent a tiny change in $x$ and $y$ respectively but what is their meaning afterwards and do we just ignore them when taking the partial derivatives?


$dx$ is a symbol which has given mathematicians trouble to define in a formal manner for a couple centuries. You can think of it as a "very small quantity", but of course this means nothing. You will see in the future that the right way to define it is by using differential forms which are elements of the cotangent space. You will introduce these notions formally in a basic course about Differential Geometry.

When you work with higher derivatives, you could formally write $dx^2$, "representing" that if $dx$ is small, then $dx^2$ is even smaller.


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