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So the problem it shows a equation that I know that they are applying the chain rule, but I just don't know exactly they applied it:

$$\frac{dv}{dx} = \frac{dv}{dt} * \frac{dt}{dx} = \frac{dv}{dt} * \frac{1}{v}$$

What are the intermarried steps involved here to get the last part of this expression?

Could someone help me understand the steps of the applied chain in this particular problem? thank you in advance

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You are interested in the change in velocity as a function of position.

Velocity is typically written as a function of time, and so is position. That is, we write $v(t)$ and $x(t)$. To get the first inequality, they simply applied the chain rule to those functions.

To get the second equality, they used the inverse function theorem combined with the chain rule. You can see the derivation of that result here.

It’s worth noting that there’s one more simplification that can be made. $\frac{dv}{dt}=a(t)$, so we have

$$\frac{dv}{dx}=a(t)+\frac{1}{v(t)}=\frac{v(t)a(t)+1}{v(t)}$$

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$dv/dx = dv/dA dA/dx$ for any A, but here you apply it with $t$.

Also, $dt/dx = 1/(dx/dt) = 1/v$.

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  • $\begingroup$ You can use \, for creating space. Or \quad for a bigger one. And \mathbf{d}x for a nicer notation. $\endgroup$
    – user373239
    Aug 30 '18 at 19:37
  • $\begingroup$ Parentheses are always worth considering in writing out the chain rule in a single line. The formatting used in the Question's body might be worth imitating for the sake of clarity. $\endgroup$
    – hardmath
    Aug 30 '18 at 20:11

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