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How can $\frac{d}{dx} \frac{dx}{dt}$ be simplified, where $x$ is a function of $t$?

My guess is that since partial derivatives commute, that $\frac{d}{dx} \frac{dx}{dt}=\frac{d}{dt}\frac{dx}{dx}=\frac{d}{dt}(1)=0$, but am not sure if I'm missing something.

Additionally would this mean that $\frac{d}{dx} \frac{dx^2}{dt}=\frac{d}{dt}\frac{dx^2}{dx}=\frac{d}{dt}(2x)=2\dot{x}$?

sbr
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  • $dx/dt=0$ already, if x is independent of t. – infinitezero Jul 22 '19 at 17:04
  • I don't know what you mean by a "solution," since no equation is present. – Paul Jul 22 '19 at 17:06
  • $\frac{d}{dx}\frac{dx}{dt}=\frac{d}{dt}$ would be the chain rule (see https://en.wikipedia.org/wiki/Chain_rule). Is $x$ a function of $t$ (i.e., as $t$ changes, does $x$ also change)? If $x$ doesn't depend on $t$, then the answer is $0$. – Axion004 Jul 22 '19 at 17:07
  • Sorry, yes. x is a function of t. And solution I suppose isn't the correct terminology here. A simplification of the expression, perhaps. – sbr Jul 22 '19 at 17:08
  • A solution requires an equation! Do you mean you want to find solutions to $\frac{d}{dx} \frac{dx}{dt} = 0$ ? – Emilio Minichiello Jul 22 '19 at 17:09
  • Maybe the OP means the second derivative? $\frac{d^2 x}{dt^2}=\frac{d}{dt}\frac{dx}{dt}$? – Axion004 Jul 22 '19 at 17:18
  • No, sorry. I've edited the question to hopefully make it a little clearer. – sbr Jul 22 '19 at 17:22
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    Unfortunately, it is still unclear what you are asking. You don't need to concentrate on commuting partial derivatives (and they don't always commute!). As written, $\frac{d}{dx}\frac{dx}{dt}=\frac{d}{dt}$ which doesn't seem helpful. Do you need to solve some sort of equation? – Axion004 Jul 22 '19 at 17:26
  • I'm not sure why everyone is asking like this is so unclear. The question (as I understand it) is just asking "How do you find the derivative of velocity with respect to position?" – mweiss Jul 22 '19 at 17:47
  • Yes, exactly. I should have presented my question that way originally, but your interpretation is correct. It was not a particularly well-posed question in the original post, so I'm glad you were able to decipher it :) – sbr Jul 22 '19 at 17:51
  • Don't confuse total derivatives with partial derivatives. Why do we type $\partial$ instead of $d$? Not for our health. – J.G. Jul 22 '19 at 17:54

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Suppose $x=f(t)$ gives the position of a particle at time $t$. Then $\frac{dx}{dt} = f'(t)$ is the velocity of the particle at time $t$. Let's write $v(t) = f'(t)$ for shorthand.

Your question is asking: What's $\frac{dv}{dx}$?

Now the issue seems to be that $v$ does not explicitly depend on $x$ at all, which would seem to imply that $\frac{dv}{dx} = 0$. This, however, oversimplifies the situation.

Am example might help: Suppose we have a particle whose position is given by $x = t^2$ for $t\ge 0$. Then $v = \frac{dx}{dt} = f'(t) = 2t$, which looks (as described above) like it doesn't depend on $x$. But appearances can be deceiving! In fact, the velocity does depend on position. Just think about the moving object: certainly, at different positions, the velocity will be different. How do we reveal this in the equation of motion?

The key is that the formula $x = t^2$ does two things: it explicitly defines the position as a function of time, but it also implicitly defines the time as a function of position. That is, if we solve for $t$, we have $t = \sqrt{x}$ (remember, we are assuming $t \ge 0$). Then substituting this into the velocity formula, we have

$$ v = 2\sqrt{x}$$

and then $\frac{dv}{dx} = \frac{1}{\sqrt{x}}$.

Now let's step away from the specific example and go back to the general case. We have $$x = f(t)$$ whose derivative with respect to $t$ is $$\frac{dx}{dt} = f'(t)$$ But also, inverting the original equation, we have $$t = f^{-1}(x)$$ which, on substituting into the equation above, gives $$\frac{dx}{dt} = f'(f^{-1}(x))$$ Now to take the derivative of this with respect to $x$, use the chain rule: $$\frac{d}{dx} \frac{dx}{dt} = f''(f^{-1}(x)) \cdot (f^{-1})'(x)$$ Finally, use the Inverse Function Theorem, which says that $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$, to conclude that $$\frac{d}{dx} \frac{dx}{dt} = f''(f^{-1}(x)) \cdot \frac{1}{f'(f^{-1}(x))}$$

Going back to our example, we have $f^{-1}(x) = \sqrt{x}$, $f'(t) = 2t$, and $f''(t) = 2$, so the formula above gives $$\frac{d}{dx} \frac{dx}{dt} = 2 \cdot \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$$ as found above.

mweiss
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