Derivatives (1pt).
Show that d(γvv)/dv = (1 − v2/c2)−3/2 for γv = 1/p1 − v2/c2.

To show that \(\frac{d(\gamma v)}{dv} = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}}\) for \(\gamma_v = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), let's proceed step by step:

1. **Express \(\gamma\):**

   \[
   \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}
   \]

2. **Multiply by \(v\) to get \(\gamma v\):**

   \[
   \gamma v = \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}}
   \]

3. **Differentiate \(\gamma v\) with respect to \(v\):**

   \[
   \frac{d}{dv} \left( \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \right)
   \]

   Use the quotient rule: \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}\), where \(u = v\) and \(v = \sqrt{1 - \frac{v^2}{c^2}}\).

4. **Find the derivatives of \(u\) and \(v\):**

   \[
   u = v \quad \Rightarrow \quad u' = 1
   \]

   \[
   v = \sqrt{1 - \frac{v^2}{c^2}} \quad \Rightarrow \quad v' = \frac{d}{dv} \left(1 - \frac{v^2}{c^2}\right)^{\frac{1}{2}}
   \]

   To differentiate \(v\), use the chain rule:

   \[
   \frac{d}{dv} \left(1 - \frac{v^2}{c^2}\right)^{\frac{1}{2}} = \frac{1}{2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} \cdot \frac{d}{dv} \left(1 - \frac{v^2}{c^2}\right)
   \]

   \[
   \frac{d}{dv} \left(1 - \frac{v^2}{c^2}\right) = -\frac{2v}{c^2}
   \]

   Therefore,

   \[
   v' = \frac{1}{2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} \cdot \left(-\frac{2v}{c^2}\right) = -\frac{v}{c^2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}
   \]

5. **Apply the quotient rule:**

   \[
   \frac{d}{dv} \left( \frac{v}{\sqrt{1 - \frac{v^2}{c^2}}} \right) = \frac{(1) \sqrt{1 - \frac{v^2}{c^2}} - v \left(-\frac{v}{c^2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}\right)}{\left( \sqrt{1 - \frac{v^2}{c^2}} \right)^2}
   \]

6. **Simplify the numerator and denominator:**

   \[
   = \frac{\sqrt{1 - \frac{v^2}{c^2}} + \frac{v^2}{c^2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}}{1 - \frac{v^2}{c^2}}
   \]

   Combine terms in the numerator:

   \[
   = \frac{\left(1 - \frac{v^2}{c^2}\right)^{\frac{1}{2}} + \frac{v^2}{c^2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}}{1 - \frac{v^2}{c^2}}
   \]

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

   \[
   = \frac{1}{\left(1 - \frac{v^2}{c^2}\right)^{\frac{3}{2}}}
   \]

Therefore,

\[
\frac{d(\gamma v)}{dv} = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}}
\]

Would you like more details or have any questions?

Here are five related questions to expand your understanding:

1. How is the Lorentz factor \(\gamma\) used in special relativity?
2. What is the physical significance of the term \(\left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}}\)?
3. Can you derive the time dilation formula using the Lorentz factor?
4. How does the velocity addition formula relate to the Lorentz transformation?
5. What are the implications of \(\gamma\) approaching infinity as \(v\) approaches \(c\)?

**Tip:** Understanding the differentiation of functions involving square roots often requires the chain rule and sometimes the quotient rule. Practice these rules to improve your skills in handling complex derivatives.

Show that d(γvv)/dv = (1 − v²/c²)⁻³/² for γv = 1/√(1 − v²/c²).

This page explains how to differentiate γv = 1/√(1 − v²/c²) and shows that d(γvv)/dv = (1 − v²/c²)⁻³/². The problem covers advanced calculus concepts such as the quotient rule and chain rule and their application to physics through the Lorentz transformation.

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