# What is the result of $\frac{d\dot x}{dx}$?

I have a problem with a step in assignment.

What is the result of $\frac{d\dot x}{dx}$ ? $x$ is the displacement, and $\dot x$ is the speed. I'm not sure if this equation itself is right.

Thank you! The problem has been solved!

While $\dot{x}$ does not vanish Chain Rule and Inverse Function Theorem can be used giving us $$\frac{d\dot{x}}{dx}=\frac{d\dot{x}}{dt}\frac{dt}{dx}=\frac{d\dot{x}}{dt}\frac{1}{\frac{dx}{dt}}=\ddot{x}\frac{1}{\dot{x}}=\frac{\ddot{x}}{\dot{x}}$$

• And where $\dot x$ does vanish, should we take it to be a limit as that point is approached? May 18 '15 at 18:29
• Yes, in such a case we should take the limit. May 18 '15 at 18:48

$$\dfrac{d\dot{x}}{dx}=\dfrac{d\dot{x}}{dt}\dfrac{dt}{dx}=\ddot{x}\dfrac{dt}{dx}=\ddot{x}(\dot{x})^{-1}=\dfrac{\ddot{x}}{\dot{x}}$$

• i.e.,$\frac{a}{v}$^^
– Mann
May 18 '15 at 18:20

Think of drawing a grpah of $x$ and a graph of $\dot x$ vs. time. Now, for each $t$, draw the curve $(x(t), \dot x(t))$ in the plane:

The slope of this curve is exactly $d\dot x / dx$: $$\frac{d\dot x}{dx}=\frac{d\dot x}{dt}\frac{dt}{dx} = \frac{a}{v}$$ This makes sense intuitively:

If there is no acceleration, the slope of the curve will zero, since $\dot x$ doesn't change with time. If on the other hand, There is no initial velocity at all, $x$ will stay the same, but $\dot x$ will change - resulting in a vertical (a.k.a. infinite) slope.

Think it like this : Lets say $x=t^n+k$ gives us the displacement of the particle where t is time What is $$\frac{dx}{dt}$$ It is the velocity right. Now taking the derivative again gives me $$\frac{d^2 x}{dt^2}$$ which is the acceleration. Now try thinking with this hint what might your question's answer will be