Questions: Find ( fracd yd t ). [ y=3 t(2 t^2-5)^5 fracd yd t=square ]

$Find ( fracd yd t ). [ y=3 t(2 t^2-5)^5 fracd yd t=square ]$

Transcript text: Find $\frac{d y}{d t}$. \[ \begin{array}{l} y=3 t\left(2 t^{2}-5\right)^{5} \\ \frac{d y}{d t}=\square \end{array} \]

Solution

Solution Steps

To find the derivative $\frac{d y}{d t}$ of the given function $y = 3t(2t^2 - 5)^5$, we will use the product rule and the chain rule. The product rule states that $(uv)' = u'v + uv'$, and the chain rule is used to differentiate composite functions.

Step 1: Define the Function

We start with the function given: \[ y = 3t(2t^2 - 5)^5 \]

Step 2: Apply the Product Rule

To find $\frac{d y}{d t}$, we use the product rule: \[ \frac{d}{dt} [u \cdot v] = u'v + uv' \] where $ u = 3t $ and $ v = (2t^2 - 5)^5 $.

Step 3: Differentiate Each Part

First, we differentiate $ u $ and $ v $: \[ u' = \frac{d}{dt} (3t) = 3 \] \[ v' = \frac{d}{dt} [(2t^2 - 5)^5] \]

To differentiate $ v $, we use the chain rule: \[ v' = 5(2t^2 - 5)^4 \cdot \frac{d}{dt} (2t^2 - 5) \] \[ \frac{d}{dt} (2t^2 - 5) = 4t \] So, \[ v' = 5(2t^2 - 5)^4 \cdot 4t = 20t(2t^2 - 5)^4 \]

Step 4: Combine Using the Product Rule

Now, we combine the results using the product rule: \[ \frac{d y}{d t} = u'v + uv' \] \[ \frac{d y}{d t} = 3(2t^2 - 5)^5 + 3t \cdot 20t(2t^2 - 5)^4 \] \[ \frac{d y}{d t} = 3(2t^2 - 5)^5 + 60t^2(2t^2 - 5)^4 \]