Questions: Given (f(t)=(5 t^2-t)(t^3-6 t^2+12)), find (f^prime(t)). (f^prime(t)=)

Transcript text: Given $f(t)=\left(5 t^{2}-t\right)\left(t^{3}-6 t^{2}+12\right)$, find $f^{\prime}(t)$. $f^{\prime}(t)=$ $\square$

Solution

Solution Steps

To find the derivative $ f'(t) $ of the function $ f(t) = (5t^2 - t)(t^3 - 6t^2 + 12) $, we will use the product rule of differentiation. The product rule states that if you have a function $ f(t) = g(t)h(t) $, then the derivative $ f'(t) $ is given by $ f'(t) = g'(t)h(t) + g(t)h'(t) $. We will first find the derivatives of $ g(t) = 5t^2 - t $ and $ h(t) = t^3 - 6t^2 + 12 $, and then apply the product rule.

Step 1: Define the Functions $ g(t) $ and $ h(t) $

Given the function $ f(t) = (5t^2 - t)(t^3 - 6t^2 + 12) $, we define: \[ g(t) = 5t^2 - t \] \[ h(t) = t^3 - 6t^2 + 12 \]

Step 2: Compute the Derivatives $ g'(t) $ and $ h'(t) $

We find the derivatives of $ g(t) $ and $ h(t) $: \[ g'(t) = \frac{d}{dt}(5t^2 - t) = 10t - 1 \] \[ h'(t) = \frac{d}{dt}(t^3 - 6t^2 + 12) = 3t^2 - 12t \]

Step 3: Apply the Product Rule

Using the product rule $ f'(t) = g'(t)h(t) + g(t)h'(t) $, we get: \[ f'(t) = (10t - 1)(t^3 - 6t^2 + 12) + (3t^2 - 12t)(5t^2 - t) \]

Step 4: Simplify the Expression

Simplifying the expression, we obtain: \[ f'(t) = 25t^4 - 124t^3 + 18t^2 + 120t - 12 \]