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Suppose a function $y(t)$. Its total derivative is:

$$ d y(t)=y'(t)dt $$

Now I want to take the total derivative again:

$$ d(dy(t))=d(y'(t)dt)=(dy'(t))dt+y'(t)d(dt)=y''(t)dt+y'(t)d(dt) $$

What is $d(dt)$ - is it zero?

Anon21
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2 Answers2

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It depends on what notation you are using. Using standard notation, where the second derivative if $\frac{d^2y}{dt^2}$, this isn't actually possible. However, if you use a more flexible notation, $d(dt) = d^2t$. In this notation, the second derivative is

$$ \frac{d^2y}{dt^2} - \frac{dy}{dt}\frac{d^2t}{dt^2} $$

This notation allows for the algebraic manipulation of higher-order differentials, which the standard notation does not allow for.

Historically, many have considered $d(dt) = 0$ iff you believe $t$ to be moving at a constant rate. This is where the origin of the more simplified version of the second derivative came from. Basically, pre-relativity, we imagined time to be flowing constantly, so $dt$ was a constant, so its differential was zero.

However, if you are just dealing with equation manipulations like you are doing, then you need to use the expanded notation.

For more information, see my paper, "Extending the Algebraic Manipulability of Differentials".

johnnyb
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You will understand an aswer if you understand for what argument do you want to take a derivative. Derivative depends on two numbers: $(d f)[x,t]=f'[x]t$. So \begin{equation}(d^2 f)[x,t]=d(f'[x]t)[x,t]=td(f'[x])[x,t]=tf''[x]t=f''[x]t^2,\end{equation} where in square brackets I wrote an argument of a function. Replacing $t$ with more convenient $dt$ you have that $(d^2 f)[x,dt]=f''[x](dt)^2$. The second derivative is a quadratic form by $dt$.

Rafael
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  • So if I have an equation such as $s^2=x^2+y^2$, I can take the total derivative twice on each side and get $(ds)^2=(dx)^2+(dy)^2$, which is the metric. is this correct? – Anon21 Mar 07 '21 at 19:51
  • I think you confuse definitions. Firstly, I wrote about second derivative of one variable function. For multivariable the second derivative is a matrix, so of what function do you take derivative? Secondly, if you curve is $s^3=x^3+y^3$, you think that metric will be $s(ds)^2=x(dx)^2+y(dy)^2$? No of course, because on a plane the metric is still the same. I can say some things about what metric is on metric space and on riemannian manifold (depends on what did you mean) and about derivatives, but I think you do not need it, so it will be better you to explain what do you want generally. – Rafael Mar 07 '21 at 20:12
  • In your example you would take the total derivative three times and get $(ds)^3=(dx)^3+(dy)^3$, which is indeed the infinitesimal version of $s^3=x^3+y^3$. – Anon21 Mar 07 '21 at 20:26
  • What is infinitesimal version?) And what if we have equation $s^{3/2}=x^{3/2}+y^{3/2}$? You will take fractional derivative?) Tell me what do you want to find and tell me in what definitions do you work, so I will try to help you. For now I see that you are just manipulating with symbols – Rafael Mar 07 '21 at 20:38
  • I am trying to construct thermodynamic cycles over an equation of state given by $S=\sqrt{X^2+X^2}$ where $S$ is the entropy and both $X$ and $Y$ are thermodynamic quantities. One usually takes the total derivative of the entropy to construct cycles, but I am left with $SdS=XdX+YdY$ if I do (since $S=\sqrt{X^2+X^2} \implies S^2=X^2+Y^2$). My idea was to take the total derivative twice to get an equation which only contains derivatives. – Anon21 Mar 07 '21 at 20:47
  • Okey, now I understand something at least. You desire cannot be solved just with taking more derivatives. And I am not sure that it is possible to write equation of state only in terms of derivatives. For example equation $dU=TdS+PdV$ as I think cannot be written in that way. And the equation of metric has few in common with taking derivatives. – Rafael Mar 07 '21 at 20:58