Differentiate the function. \( s(t)=6 t-2\left(3 t^{4}+4\right)^{3} \) \( s^{\prime}(t)=\square \)

1. Write the function: \[ s(t)=6t-2\left(3t^4+4\right)^3. \] 2. Differentiate \(6t\): \[ \frac{d}{dt}\left(6t\right)=6. \] 3. Differentiate \(-2\left(3t^4+4\right)^3\) using the chain rule: - Let \( u = 3t^4+4 \). Then the term becomes \(-2u^3\). - Differentiate \( u^3 \) with respect to \( u \): \[ \frac{d}{du}\left(u^3\right)=3u^2. \] - Differentiate \( u \) with respect to \( t \): \[ \frac{du}{dt}=\frac{d}{dt}\left(3t^4+4\right)=12t^3. \] - Apply the chain rule: \[ \frac{d}{dt}\left(-2u^3\right)=-2\cdot 3u^2\cdot \frac{du}{dt}=-6u^2\cdot 12t^3=-72t^3u^2. \] - Substitute back \( u=3t^4+4 \): \[ -72t^3\left(3t^4+4\right)^2. \] 4. Combine the derivatives: \[ s'(t)=6-72t^3\left(3t^4+4\right)^2. \] Thus, the derivative is: \[ s'(t)=6-72t^3\left(3t^4+4\right)^2. \]