Find the derivative of the function using the definition of derivative. \[ \begin{array}{l}f^{\prime}(t)=5.5 t^{2}+3 t \\ \text { State the domain of the function. (Enter your answer using interval notation.) } \\ \text { State the domain of its derivative. (Enter your answer using interval notation.) } \\ \text { Need Help? }\end{array} \text { Read lt Watchlt } \]

To find the derivative of the function \( f(t) = 5.5t^2 + 3t \) using the definition of the derivative, we will use the formula: \[ f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \] Given: \[ f(t) = 5.5t^2 + 3t \] Let's calculate the derivative step by step. Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{h\rightarrow 0}\left(\frac{\left(5.5\left(t+h\right)^{2}+3\left(t+h\right)\right)-\left(5.5t^{2}+3t\right)}{h}\right)\) - step1: Add the terms: \(\lim _{h\rightarrow 0}\left(\frac{\left(5.5t^{2}+11th+5.5h^{2}+3t+3h\right)-\left(5.5t^{2}+3t\right)}{h}\right)\) - step2: Remove the parentheses: \(\lim _{h\rightarrow 0}\left(\frac{5.5t^{2}+11th+5.5h^{2}+3t+3h-\left(5.5t^{2}+3t\right)}{h}\right)\) - step3: Subtract the terms: \(\lim _{h\rightarrow 0}\left(\frac{11th+5.5h^{2}+3h}{h}\right)\) - step4: Reduce the fraction: \(\lim _{h\rightarrow 0}\left(11t+5.5h+3\right)\) - step5: Rewrite the expression: \(\lim _{h\rightarrow 0}\left(11t\right)+\lim _{h\rightarrow 0}\left(5.5h\right)+\lim _{h\rightarrow 0}\left(3\right)\) - step6: Calculate: \(11t+0+3\) - step7: Calculate: \(11t+3\) The derivative of the function \( f(t) = 5.5t^2 + 3t \) using the definition of the derivative is \( f'(t) = 11t + 3 \). Now, let's determine the domain of the function and its derivative. The domain of the function \( f(t) = 5.5t^2 + 3t \) is all real numbers, which can be represented in interval notation as \( (-\infty, \infty) \). The domain of the derivative \( f'(t) = 11t + 3 \) is also all real numbers, which can be represented in interval notation as \( (-\infty, \infty) \).