1. **State the problem:** Find the first and second derivatives of the function $$T(v) = \frac{d(1+k(v+w)^2)}{v}$$ with respect to $$v$$. 2. **Rewrite the function:** $$T(v) = \frac{d(1+k(v+w)^2)}{v} = d \cdot \frac{1+k(v+w)^2}{v}$$ 3. **Use the quotient rule for differentiation:** If $$f(v) = \frac{g(v)}{h(v)}$$, then $$f'(v) = \frac{g'(v)h(v) - g(v)h'(v)}{h(v)^2}$$ Here, $$g(v) = d(1+k(v+w)^2)$$ and $$h(v) = v$$. 4. **Calculate derivatives of numerator and denominator:** - $$g'(v) = d \cdot k \cdot 2(v+w) = 2dk(v+w)$$ - $$h'(v) = 1$$ 5. **Apply quotient rule:** $$T'(v) = \frac{2dk(v+w) \cdot v - d(1+k(v+w)^2) \cdot 1}{v^2}$$ 6. **Simplify numerator:** $$2dkv(v+w) - d(1+k(v+w)^2) = d\left(2kv(v+w) - 1 - k(v+w)^2\right)$$ 7. **Rewrite numerator inside brackets:** $$2kv(v+w) - k(v+w)^2 - 1 = k\left(2v(v+w) - (v+w)^2\right) - 1$$ 8. **Expand terms inside brackets:** $$2v(v+w) = 2v^2 + 2vw$$ $$(v+w)^2 = v^2 + 2vw + w^2$$ 9. **Calculate:** $$2v^2 + 2vw - (v^2 + 2vw + w^2) = 2v^2 + 2vw - v^2 - 2vw - w^2 = v^2 - w^2$$ 10. **Substitute back:** $$k(v^2 - w^2) - 1$$ 11. **Final expression for first derivative:** $$T'(v) = \frac{d\left(k(v^2 - w^2) - 1\right)}{v^2}$$ 12. **Find second derivative:** Rewrite first derivative as $$T'(v) = d \cdot \frac{k(v^2 - w^2) - 1}{v^2} = d \cdot \left(k \frac{v^2 - w^2}{v^2} - \frac{1}{v^2}\right)$$ 13. **Simplify inside:** $$\frac{v^2 - w^2}{v^2} = 1 - \frac{w^2}{v^2}$$ So, $$T'(v) = d \left(k \left(1 - \frac{w^2}{v^2}\right) - \frac{1}{v^2}\right) = d \left(k - k \frac{w^2}{v^2} - \frac{1}{v^2}\right)$$ 14. **Differentiate term by term:** - Derivative of $$k$$ is 0 (constant) - Derivative of $$-k w^2 v^{-2}$$ is $$-k w^2 \cdot (-2) v^{-3} = 2 k w^2 v^{-3}$$ - Derivative of $$-v^{-2}$$ is $$-(-2) v^{-3} = 2 v^{-3}$$ 15. **Sum derivatives:** $$T''(v) = d \left(2 k w^2 v^{-3} + 2 v^{-3}\right) = 2 d v^{-3} (k w^2 + 1)$$ 16. **Final answer:** $$\boxed{T'(v) = \frac{d\left(k(v^2 - w^2) - 1\right)}{v^2}}$$ $$\boxed{T''(v) = \frac{2 d (k w^2 + 1)}{v^3}}$$