Find the derivative of the function. \[ f(t)=t \cos (t)+t^{2} \sin (t) \]

To find the derivative of the function \( f(t) = t \cos(t) + t^2 \sin(t) \), we'll apply the product rule, which states that if you have a function in the form of \( u(t) v(t) \), then its derivative is \( u'(t)v(t) + u(t)v'(t) \). 1. For the first term \( t \cos(t) \): - Let \( u = t \) and \( v = \cos(t) \). - Thus, \( u' = 1 \) and \( v' = -\sin(t) \). - The derivative is: \[ (u' v + u v') = (1 \cdot \cos(t) + t \cdot (-\sin(t))) = \cos(t) - t \sin(t). \] 2. For the second term \( t^2 \sin(t) \): - Let \( u = t^2 \) and \( v = \sin(t) \). - Thus, \( u' = 2t \) and \( v' = \cos(t) \). - The derivative is: \[ (u' v + u v') = (2t \cdot \sin(t) + t^2 \cdot \cos(t)). \] Now, combine the derivatives from both terms: \[ f'(t) = \left( \cos(t) - t \sin(t) \right) + \left( 2t \sin(t) + t^2 \cos(t) \right). \] Combining like terms yields: \[ f'(t) = \cos(t) + (2t - t) \sin(t) + t^2 \cos(t) = \cos(t) + t \sin(t) + t^2 \cos(t). \] Finally, simplify the expression: \[ f'(t) = (1 + t^2) \cos(t) + t \sin(t). \] Thus, the derivative of the function is: \[ f'(t) = (1 + t^2) \cos(t) + t \sin(t). \]