The function \(g(t) = t^2 e^t\) is differentiated. What is \(g'(t)\)?
["Understanding the Differentiation of ( g(t) = t^2 e^t ): Finding ( g'(t) )", "Calculus is essential for analyzing how functions change, and one of the most powerful tools in differential calculus is the differentiation of products. A common application involves functions expressed as the product of two parts—such as a polynomial and an exponential function—making ( g(t) = t^2 e^t ) a classic example.", "### What is ( g(t) )?", "Given:\n[\ng(t) = t^2 e^t\n]\nHere, ( g(t) ) combines a quadratic polynomial ( t^2 ) with the exponential function ( e^t ). Because these two components multiply, their derivatives interact, requiring the product rule for differentiation.", "### Applying the Product Rule", "The product rule states that if ( g(t) = u(t) \cdot v(t) ), then:\n[\ng'(t) = u'(t) \cdot v(t) + u(t) \cdot v'(t)\n]", "For our function:\n- Let ( u(t) = t^2 ) → ( u'(t) = 2t )\n- Let ( v(t) = e^t ) → ( v'(t) = e^t )", "Now apply the rule:\n[\ng'(t) = (2t) \cdot e^t + t^2 \cdot (e^t)\n]", "### Simplify the Derivative", "Factoring out ( e^t ):\n[\ng'(t) = e^t (2t + t^2)\n]", "Alternatively, writing in standard polynomial order:\n[\ng'(t) = t^2 e^t + 2t e^t\n]", "### Why This Matters in Applications", "Differentiating expressions like ( t^2 e^t ) is crucial in fields such as physics, engineering, and economics, where growth processes with variable rates are modeled. The resulting derivative reveals instantaneous rate of change, informing optimization, modeling, and system analysis.", "### Final Answer", "[\n\boxed{g'(t) = e^t (t^2 + 2t)}\n]", "Understanding how to differentiate products like ( g(t) = t^2 e^t ) enables deeper insights into complex dynamic systems governed by exponential and polynomial behaviors. Mastering this technique strengthens your toolkit for tackling advanced calculus problems and real-world modeling challenges."]
