Find the derivative of the function. \( s(t)=\frac{5}{t^{2}+3 t-1} \)

Evaluate the derivative by following steps: - step0: Solution: \(\frac{d}{dt}\left(\frac{5}{t^{2}+3t-1}\right)\) - step1: Use differentiation rules: \(5\times \frac{d}{dt}\left(\frac{1}{t^{2}+3t-1}\right)\) - step2: Rewrite the expression: \(5\times \frac{d}{dt}\left(\left(t^{2}+3t-1\right)^{-1}\right)\) - step3: Calculate the derivative: \(5\left(-\frac{2t+3}{\left(t^{2}+3t-1\right)^{2}}\right)\) - step4: Multiply the terms: \(-\frac{10t+15}{\left(t^{2}+3t-1\right)^{2}}\) - step5: Evaluate the power: \(-\frac{10t+15}{t^{4}+7t^{2}+1+6t^{3}-6t}\) To find the derivative of the function \( s(t) = \frac{5}{t^{2} + 3t - 1} \), we can use the quotient rule or rewrite it in a form suitable for differentiation. Here, I will use the rewritten form: 1. Rewrite the function: \[ s(t) = 5(t^{2} + 3t - 1)^{-1} \] 2. Apply the chain rule: The derivative of \( s(t) \) can be calculated as follows: \[ s'(t) = 5 \cdot (-1)(t^{2} + 3t - 1)^{-2} \cdot (2t + 3) \] 3. Simplifying the expression: \[ s'(t) = -\frac{5(2t + 3)}{(t^{2} + 3t - 1)^{2}} \] 4. Further simplification leads to: \[ s'(t) = -\frac{10t + 15}{(t^{2} + 3t - 1)^{2}} \] Thus, the derivative of the function \( s(t) \) is: \[ s'(t) = -\frac{10t + 15}{(t^{2} + 3t - 1)^{2}} \]