1. **State the problem:** We are given the mass formula from special relativity: $$m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$ where $m_0$ and $c$ are constants, and $v$ is the velocity. We need to find the derivative $\frac{dm}{dv}$. 2. **Rewrite the formula for easier differentiation:** Express $m$ as: $$m = m_0 \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}}$$ 3. **Apply the chain rule:** Since $m_0$ and $c$ are constants, differentiate using the power rule and chain rule only: $$\frac{dm}{dv} = m_0 \cdot \left(-\frac{1}{2}\right) \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}} \cdot \frac{d}{dv} \left(1 - \frac{v^2}{c^2}\right)$$ 4. **Differentiate the inner function:** $$\frac{d}{dv} \left(1 - \frac{v^2}{c^2}\right) = 0 - \frac{2v}{c^2} = -\frac{2v}{c^2}$$ 5. **Substitute back:** $$\frac{dm}{dv} = m_0 \cdot \left(-\frac{1}{2}\right) \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}} \cdot \left(-\frac{2v}{c^2}\right)$$ 6. **Simplify the expression:** $$\frac{dm}{dv} = m_0 \cdot \cancel{-\frac{1}{2}} \cdot \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}} \cdot \cancel{-\frac{2v}{c^2}} = m_0 \cdot \frac{v}{c^2} \left(1 - \frac{v^2}{c^2}\right)^{-\frac{3}{2}}$$ 7. **Final answer:** $$\boxed{\frac{dm}{dv} = \frac{m_0 v}{c^2 \left(1 - \frac{v^2}{c^2}\right)^{\frac{3}{2}}}}$$ This formula shows how the mass changes with velocity $v$ as it approaches the speed of light $c$.