This doesn't seem to be working? Relativity

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The discussion focuses on evaluating the derivative of the force equation \(\vec{F} = \frac{d}{dt} \frac{m \vec{v}}{\sqrt{1 - v^2/c^2}}\) to determine the particle's acceleration. The user attempted to apply the quotient rule while incorrectly treating mass and the denominator as constants, leading to incorrect results. The community emphasizes that \(v^2\) cannot be treated as a constant during differentiation. A reminder is provided that the relationship \(\frac{dv}{dt} = \frac{\vec{a} \cdot \vec{v}}{v}\) should be considered in the evaluation. Understanding the correct application of differentiation rules is crucial for solving this problem accurately.
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Evaluate the derivative of \vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}} to find the acceleration a = \frac {dv}{dt} of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}
\frac {F}{m} (1 + \frac {v^2}{3c^2})
 
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IntegrateMe said:
Evaluate the derivative of \vec{F} = \frac {d} {dt} \frac {m \vec{v}} {\sqrt{1 - v^2/c^2}} to find the acceleration a = \frac {dv}{dt} of the particle.

So, basically, I just tried to use the quotient rule and treat m, and the whole bottom of the fraction as constants. I didn't end up getting the right answer and I can't figure out why. As reference, here are two answers I've tried, both wrong:

\frac {F}{m} \sqrt {1 - \frac {v^2}{c^2}}
\frac {F}{m} (1 + \frac {v^2}{3c^2})
How can you treat v2 as a constant?
 
Remember the fact that:

\frac{dv}{dt}=\frac{\vec{a}\cdot \vec{v}}{v}
 
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