To find the derivative of \(t^2 \sin(3t - 5)\) with respect to \(t\), we'll apply the product rule of differentiation.

The product rule states that if \(u\) and \(v\) are differentiable functions of \(t\), then the derivative of their product \(uv\) is given by:

\[(uv)' = u'v + uv'\]

In this case, \(u = t^2\) and \(v = \sin(3t - 5)\).

First, find the derivatives of \(u\) and \(v\) with respect to \(t\):

\(u' = 2t\)

\(v' = 3 \cos(3t - 5)\) (derivative of \(\sin(3t - 5)\) with respect to \(t\) using chain rule)

Now, apply the product rule:

\[(t^2 \sin(3t - 5))' = (2t) \sin(3t - 5) + (t^2) (3 \cos(3t - 5))\]

Simplify:

\[2t \sin(3t - 5) + 3t^2 \cos(3t - 5)\]

So, the derivative of \(t^2 \sin(3t - 5)\) with respect to \(t\) is \(2t \sin(3t - 5) + 3t^2 \cos(3t - 5)\)