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Let us have: $$\cases{ F = F(x,y)=const \\ y = y(x)} $$ I am taking a second differential over function $F$, i.e. $d^2F$. During calculations I have this term: $$\frac{d^2x}{dx^2}$$ I feel awkward, as I want it to be zero, but the only way I can see it is if I rewrite it like this: $$\frac{d^2x}{dx^2} = \frac{d}{dx}\left(\frac{dx}{dx}\right)$$ But I also doubt if that is valid..

positive
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$\frac {d^{2} x}{dx^{2}}=\frac d{dx} ({\frac d {dx} (x)})=\frac d {dx} (1)=0$. This is quite valid.

Kavi Rama Murthy
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    Am I overthinking it? Is it valid to apply a differential together with $dx$ as derivative over $x$ first and then apply second differential with the $dx$ ? I.e. is it okay to change the order? I guess once we have $d^2f(x)$ we can not put $dx$ under differential so easily. – positive Oct 04 '19 at 08:21
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    @positive You might be confused by the notation. $\text{d}^2 f/\text{d} x^2$ is not a fraction where you first compute the numerator and then divide it by the denominator. Rather, as in the answer, it is the square of the operator $(\text{d}/\text{d}x)$, applied to the function $f$, i.e. you differentiate once, and the differentiate the result again. – Toffomat Oct 04 '19 at 08:28
  • okay... Because when I was arriving to this $\frac{d^2x}{dx^2}$ expression, I treated $dx^2$ as a separate part as I was diving bunch of elements with $dx^2$, i.e. as denumerator, but now I have to treat it like a part of the operator.. @Toffomat, could you please tell where do I have this confusion from? – positive Oct 04 '19 at 08:33
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    @positive Well, the first derivative is defined as the limit of differences ($\Delta f/\Delta x$), and so the operator is written as $\text d /\text d x$. However, the operator is not a fraction, and it is best to think of it as a single object that maps functions to their derivatives. – Toffomat Oct 04 '19 at 08:51