# Problem in Kinetic enery and relativistic force

In my textbook there's the definition of the kinetic energy in analogy with classical mechanics

$$dT=dW= \bf{F} \cdot d \bf{x}$$ where dW is the infinitesimal work done by the force $\bf{F}= \frac{d}{dt}(m \gamma v)$

After that it states

$$\frac{dT}{dt}= \bf{F} \cdot \frac{d\bf{x}}{dt}$$ while I think it should be $$\frac{dT}{dt}= \bf{F} \cdot \frac{d\bf{x}}{dt} + \frac{dF}{dt} \cdot d\bf{x}$$

Am I missing something here?

• Hint: Consider a relation of the form $dy = f dx$. What should $dy/dx$ be? – Raziman T V Jan 13 '17 at 11:57
• no, Run Like Hell is right in the general mathematics of calculus, it must just be that the force is not time varying $dF/dt=0$, or is small and can be neglected. It's a bit difficult without the context to validate that assumption. What is your $dT/dt$, (rate of change of k.e.) used for? – JMLCarter Jan 13 '17 at 12:18
• We use the $dT$ to find $T$ after an integration. It writes down that derivative in an explicit way and then integrates to find $T=mc^2(\gamma -1)$ – Run like hell Jan 13 '17 at 13:53

You just devide $dT=Fdx$ by $dt$ to get the result.