Question

Find the derivative of the function. \(g(t)=t^{2} 2^{t}\)

Short answer

The derivative of the function, \(g(t)\), is \(g'(t) = t2^{t}(2 + t \ln(2))\).

Step-by-step solution

Identify the two functions and apply the product rule

Identify the two functions. Here, we have \(f(t) = t^{2}\) and \(h(t) = 2^{t}\). The product rule states that if you have two functions multiplied together (i.e., \(g(t) = f(t)h(t)\)), then the derivative, \(g'(t)\), is given by \(f'(t)h(t) + f(t)h'(t)\). In this case, \(f'(t)\) is the derivative of \(t^{2}\), which is \(2t\), and we'll solve for \(h'(t)\) in the next step.

Differentiate \(2^{t}\) using the chain rule

The chain rule of differentiation states that the derivative of a composition of functions is the derivative of the outside function evaluated at the inside function, multiplied by the derivative of the inside function. In this case, the 'outside' function is \(e^u\) and the 'inside' function is \(u= t \ln(2)\). The derivative \(h'(t)\) is then \(h'(t) = \ln(2)2^{t}\).

Apply the product rule

Using the product rule, the derivative \(g'(t)\) is \(f'(t)h(t) + f(t)h'(t) = 2t \cdot 2^{t} + t^{2} \cdot \ln(2)2^{t}\).

Simplify the result

The final step is to factor out common factors for simplification. Factoring out \(t2^{t}\), we get the final derivative: \(g'(t) = t2^{t}(2+t \ln(2))\).