Questions: Find the derivative of the function, s(t)=(6 t^3-t)(t^3-3 t+2) s^prime(t)=

Transcript text: Find the derivative of the function, \[ \begin{array}{l} s(t)=\left(6 t^{3}-t\right)\left(t^{3}-3 t+2\right) \\ s^{\prime}(t)= \end{array} \]

Solution

Solution Steps

Step 1: Define the Functions

We start with the function \( s(t) = (6t^3 - t)(t^3 - 3t + 2) \). We identify the two components of the product:

\( u(t) = 6t^3 - t \)
\( v(t) = t^3 - 3t + 2 \)

Step 2: Calculate the Derivatives

Next, we compute the derivatives of \( u(t) \) and \( v(t) \):

The derivative \( u'(t) = 18t^2 - 1 \)
The derivative \( v'(t) = 3t^2 - 3 \)

Step 3: Apply the Product Rule

Using the product rule, we find the derivative of \( s(t) \): \[ s'(t) = u'(t)v(t) + u(t)v'(t) \] Substituting the derivatives and the original functions, we have: \[ s'(t) = (18t^2 - 1)(t^3 - 3t + 2) + (6t^3 - t)(3t^2 - 3) \]