Relativistic Kinetic Energy Derivation

[RedDelicious]

Homework Statement

This problem comes from an intermediate step in the textbook's derivation of relativistic energy. It states that

$$E_k\:=\:\int _0^u\frac{d\left(\gamma mu\right)}{dt}dx$$

then leaves the following intermediate calculation as an exercise to the reader:

Show that

$$d\left(\gamma mu\right)=\frac{m}{\left(1-\frac{u^2}{c^2}\right)^{\frac{3}{2}}\:}du$$

Homework Equations

Where

$$\gamma \:=\:\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}$$

The Attempt at a Solution

:
[/B]

I tried using the product rule

$$\frac{d}{du}\left(\gamma mu\right)=m\:\frac{d}{du}\left(\gamma u\right)=\:m\left(\gamma 'u+u'\gamma \right)=\:m\left[u\:\frac{d}{du}\left(1-\frac{u^2}{c^2}\right)^{-\frac{1}{2}}+\left(1-\frac{u^2}{c^2}\right)^{-\frac{1}{2}}\right]$$

$$=m\left[\frac{u^2}{c^2}\left(1-\frac{u^2}{c^2}\right)^{-\frac{3}{2}}+\left(1+\frac{u^2}{c^2}\right)^{-\frac{1}{2}}\right]$$

Anyone know what I'm doing wrong?

 

[TSny]

Everything looks good so far. Keep going.

[Oops, I do see a typo error in your last equation. Check the signs inside the parentheses.]

 

[RedDelicious]

Everything looks good so far. Keep going.

[Oops, I do see a typo error in your last equation. Check the signs inside the parentheses.]

Thank you!

I figured it out. I can't believe I did not see that this entire time.