$v^{i}= \frac{dx^{i}}{dt}=c\frac{dx^{i}}{d\tau}\left( \frac{dx^{0}}{d\tau} \right)^{-1}$ I do not understand where the last two equalities emerge

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I'm considering a parametrization of the worldline $ \tau \longrightarrow x^{\mu}(\tau).$

Considering the projection of the universe line in three-dimensional space, how do speed components in three-dimensional space be: $$v^{i}= \frac{dx^{i}}{dt}=c\frac{dx^{i}}{d\tau}\left( \frac{dx^{0}}{d\tau} \right)^{-1}$$ I do not understand where the last two equalities emerge.

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Keep in mind it is convention that English superscripts are for spatial dimensions while $0$ superscript is for the time dimension and Greek superscripts are all dimensions inclusive. Velocity is defined to be the change of position per unit time. $$v^i=\frac{dx^i}{dt}$$ The time coordinate is $x^0=ct$. Introduce a new coordinate $\tau$, sometimes referred to as the "proper time", and apply chain rule. $$\frac{dx^i}{dt}=\frac{dx^i}{d\tau}\frac{d\tau}{dt}=\frac{dx^i}{d\tau}\left(\frac{dt}{d\tau}\right)^{-1}=\frac{dx^i}{d\tau}\left(\frac{d(x^0/c)}{d\tau}\right)^{-1}=c\frac{dx^i}{d\tau}\left(\frac{dx^0}{d\tau}\right)^{-1}$$