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Can anyone kindly explain to me that if $\frac{d}{dt}(x) = x^{'}$ then what's $\frac{d}{dx}(x^{'}) $ and $\frac{d}{dx^{'}}(x)$?

Esan
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    There is a similar question over on the physics stackexchange where they explain why the two expressions are 0. – BeaumontTaz Sep 10 '14 at 08:25
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    At times I have seen manipulations like these: $$\frac{d \dot x}{dx} = \frac{d}{dx} \left(\frac{dx}{dt}\right) = \frac{d}{dt} \left(\frac{dx}{dx}\right)= \frac{d}{dt} 1 = 0 \tag{1}$$

    $$\frac{dx}{d\dot{x}} = \left(\frac{dx}{dt} \right) \div \left(\frac{d \dot x}{dt} \right) = \frac{\dot x}{\ddot x} \tag{2}$$

    $$\dot x \frac{d \dot x}{dx} = \left(\frac{dx}{dt} \right) \left(\frac{d \dot x}{dx} \right) = \frac{d\dot x}{dt} =\ddot x \tag{3}$$

    If such jugglery is possible, among other things, this begs the question as to why $(1)$ and $(2)$ are not reciprocals.

     – Macavity Sep 10 '14 at 10:39

1 Answers1

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Let me assume x = x(t) , hence the velocity can be determined as mentioned above $$ \frac{d x}{dt}= x'$$, suppose x(t) is of class $C^k$ where k $\geq$ 2. therefore atleast higher derivatives, upto order 2, of x exists and continuous everywhere. The derivatives can be represented as below $$x'=x'(t)$$ $$x''=x''(t)$$.
Your question : Chain rule method $$(\frac{d}{dx})x'=\frac{dx'}{dt}\frac{dt}{dx}$$ simplify, algebraically $$(\frac{d}{dx})x'=\frac{x''}{x'}$$ similarly, solution to next one is arrived.$$ (\frac{d}{dx'})x =\frac{x'}{x''}$$

selva kumar
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    I think it's useful to point out that the second-to-last equation implies $\dfrac{d}{dx}(x')^2=2x'\dfrac{dx'}{dx}=2x''$. If we write this in terms of the velocity $v$ and acceleration $a$, this becomes $\dfrac{d}{dx}v^2=2a$. If $a$ is constant, we may integrate to obtain $v_1^2-v_0^2=2a(x_1-x_0)$ which is one of the standard kinematic equations. So while $\dfrac{dx'}{dx}$ may look unusual, it really represents something more familiar. – Semiclassical Sep 11 '14 at 01:12
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    From the discussion I believe, if velocity is constant, in which case x would be linear function of time, then $\frac{d}{dx}(x^{'}) = 0.$ On the other hand, if velocity is not constant, in which case x would be nonlinear function of time, then $\frac{d}{dx}(x^{'})$ would be as @selva kumar computed. – Esan Sep 12 '14 at 07:36