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** for differentiation, which states that if you have a function \( u(t) \cdot v(t) \), its derivative is: \[ \frac{d}{dt}[u(t) \cdot v(t)] = u'(t) \cdot v(t) + u(t) \cdot v'(t). \] Let's differentiate each term of \( f(t) \) separately. ### 1. Differentiating \( t \cos(t) \): Let \( u(t) = t \) and \( v(t) = \cos(t) \). - \( u'(t) = 1 \) - \( v'(t) = -\sin(t) \) Applying the product rule: \[ \frac{d}{dt}[t \cos(t)] = 1 \cdot \cos(t) + t \cdot (-\sin(t)) = \cos(t) - t \sin(t) \] ### 2. Differentiating \( t^{2} \sin(t) \): Let \( u(t) = t^{2} \) and \( v(t) = \sin(t) \). - \( u'(t) = 2t \) - \( v'(t) = \cos(t) \) Applying the product rule: \[ \frac{d}{dt}[t^{2} \sin(t)] = 2t \cdot \sin(t) + t^{2} \cdot \cos(t) = 2t \sin(t) + t^{2} \cos(t) \] ### Combining the Results: Add the derivatives of both terms to get the derivative of \( f(t) \): \[ f'(t) = (\cos(t) - t \sin(t)) + (2t \sin(t) + t^{2} \cos(t)) \] Simplifying: \[ f'(t) = \cos(t) + t \sin(t) + t^{2} \cos(t) \] This can be factored for a more compact expression: \[ f'(t) = (1 + t^{2}) \cos(t) + t \sin(t) \] ### Final Answer: \[ f\,'(t) = (1 + t^{2})\,\cos t\;+\;t\,\sin t \]