# How to derive the law of velocity transformation using chain rule?

I know there is a straightforward way to derive the law of velocity transformation in Special Relativity by just dividing $dx$ on $dt$. But I am interested here to apply the chain rule and see how to obtain the same formula. I just begin by writing the differential of $dx$ with respect to other coordinates; $$\frac{dx}{dt}=\frac{\partial x}{\partial x'}\frac{\partial x'}{\partial t}+\frac{\partial x}{\partial t'}\frac{\partial t'}{\partial t}$$ But $\frac{\partial x'}{\partial t}=\frac{\partial x'}{\partial t'}\frac{\partial t'}{\partial t}$

So, $$\frac{dx}{dt}=\frac{\partial x}{\partial x'}\frac{\partial x'}{\partial t'}\frac{\partial t'}{\partial t}+\frac{\partial x}{\partial t'}\frac{\partial t'}{\partial t}$$ Substitute $u$ for $\frac{dx}{dt}$ and $u'$ for $\frac{\partial x'}{\partial t'}$ Also, from the two equation of LT $\frac{\partial x}{\partial x'}=\gamma$, $\frac{\partial t'}{\partial t}=\frac{1}{\gamma}$, $\frac{\partial x}{\partial t'}=v\gamma$ and $\frac{\partial t'}{\partial t}=\frac{1}{\gamma}$ (here, I did not use the inverse LT to substitute for $\frac{\partial t'}{\partial t}$, instead I used the main transformation) yields; $$u=\gamma u'\frac{1}{\gamma}+v\gamma\frac{1}{\gamma}=u'+v$$ However, this is non-relativistic transformation of the velocities. So probably, I missed something in the chain rule or in the substitution. Regards,

• I don't know the answer (yet), but I suspect it has something to do with mixing partial and total derivatives, and being careful about what's kept constant. I'll see if I can work it out. Jul 16 '17 at 14:40

• Great, I got the same result if dxdt=dxdx′dxdt′dtdt..Still I dont get why the first equation should be in total derivative form. I mean this is just the second term of my original equation in partial derivative. Jul 16 '17 at 21:49