Relativistic Momentum derivative respect to velocity

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SUMMARY

The discussion centers on the derivative of relativistic momentum, specifically the expression d(γmu) = m(1-(u^2/c^2))^(-3/2). The variable γ is defined as γ = (1-(u^2/c^2))^(-1/2). The user attempts to derive the momentum's derivative with respect to velocity (u) using the chain rule but encounters an error related to the factor (u/c^2). The correct approach involves applying both the product rule and the chain rule to accurately compute the derivative.

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Curtis15
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Homework Statement



show the d(γmu) = m(1-(u^2/c^2))^(-3/2)

Homework Equations



C = constant, m = constant, γ= (1-(u^2/c^2))^(-1/2)

The Attempt at a Solution



So in calculating d(γmu)/du, I thought I would take out the m first since it is a constant and then just add it in later.

m * d(γu)/du

So I just have to calculate what γ is. γ= (1-(u^2/c^2))^(-1/2), so by chain rule the derivative of this is (-1/2)((1-(u^2/c^2))^(-3/2) * -2u/c^2. Then I just multiply this by m and I get
m(u/c^2)(1-(u^2/c^2))^(-3/2).

This differs from the correct solution by the factor (u/c^2) which i got from doing the chain rule and calculating the derivative of the inside with respect to u. If someone could please show me the error in my ways, that would be greatly appreciated. Thank you.
 
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Is it \gamma * m * u ? If so, you must use the product rule first, then the chain rule.
 

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