Differentiating x2 w.r.t Time: Must I Differentiate \dot x?

AI Thread Summary
The discussion focuses on differentiating the time-dependent variable x squared with respect to time t. The original poster correctly applies the chain rule but questions whether they need to differentiate \dot x, which represents the derivative of x with respect to time. Responses clarify that \dot x does not need to be differentiated further in this context, and suggest clearer notation to avoid ambiguity. The proper expression for the derivative is emphasized as \dot{x} \frac{dx^2}{dx} or \left(\frac{dx}{dt}\right)\left(\frac{dx^2}{dx}\right). Ultimately, the correct derivative of x squared with respect to x is simply 2x.
Niles
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Homework Statement


Hi all.

I have a time-dependent variable called x. I wish to differentiate x2 with respect to the time t, and this is what I have done:

<br /> \frac{d}{dt}x^2 = \frac{d}{dx}x^2\frac{dx}{dt} = \frac{d}{dx}x^2\dot x.<br />

where the dot over x denotes differentiation with respect to time t. Now my question is, must I differentiate \dot x also with respect to x, since it is standing on its right side?
 
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Nah you dont. It is a little bit ambiguous how you have written it, but in this context it is just d(x^2)/dx MULTIPLED by x dot.
 
Last edited:
How is the proper way to write it then?

Thanks for replying.
 
Its not that how you have written it is wrong, it is just that it could possibly be misinterpreted. I personally would have witten it with the x dot on the other side of the derivative operator just so there will be no confusion:

<br /> <br /> \dot x \frac{d}{dx} x^2<br /> <br />
 
Better would be
\dot{x}\frac{dx^2}{dx}
or just
\left(\frac{dx}{dt}\right)\left(\frac{dx^2}{dx}\right)

Now, what is
\frac{d x^2}{dx}?
 
HallsofIvy said:
Now, what is
\frac{d x^2}{dx}?

Ahh, now you are just teasing me :smile:

Thanks to all for helping.
 
I don't think Halls was teasing - you shouldn't leave it as dx^2/dx. I think you understand, but I'm not 100% certain.
 
The reason why I didn't evaluate it in my first post was because I wasn't sure if I had to differentiate \dot x as well, but now I would of course just write 2x instead.
 
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