Find \( \frac{d s}{d t} \) if \( s=\frac{t}{7 t+3} \) \( \frac{d s}{d t}=\square \)

We are given \[ s = \frac{t}{7t+3}. \] To find \(\frac{ds}{dt}\), we use the quotient rule. The quotient rule states that if \[ s = \frac{u(t)}{v(t)}, \] then \[ \frac{ds}{dt} = \frac{v(t) \cdot \frac{d}{dt}[u(t)] - u(t) \cdot \frac{d}{dt}[v(t)]}{[v(t)]^2}. \] Let \[ u(t) = t \quad \text{and} \quad v(t) = 7t+3. \] First, compute the derivatives: \[ \frac{d}{dt}[u(t)] = 1 \quad \text{and} \quad \frac{d}{dt}[v(t)] = 7. \] Now substitute into the quotient rule formula: \[ \frac{ds}{dt} = \frac{(7t+3)(1) - t(7)}{(7t+3)^2}. \] Simplify the numerator: \[ (7t+3) - 7t = 3. \] Thus, we have: \[ \frac{ds}{dt} = \frac{3}{(7t+3)^2}. \] So, the derivative is: \[ \frac{ds}{dt} = \boxed{\frac{3}{(7t+3)^2}}. \]