Question

Find the derivative of the function using the definition of the derivative. State the domain of the function and the domain of the derivative.

25. \(A\left( p \right) = 4{p^3} + 3p\)

Short answer

The derivative of the function is \(12{p^2} + 3\). The domain of the function \(f\) and derivative \(f'\) is \(\mathbb{R}\).

Step-by-step solution

Condition for derivative of the function

The derivative of the function \(f\) at any number \(x\):

\(f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\)

The function \(f'\) is called the derivative of the function.

Determine the derivative of the function and domain of a function

Evaluate the derivative of the function as shown below:

\(\begin{aligned}A'\left( p \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{A\left( {p + h} \right) - A\left( p \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {4{{\left( {p + h} \right)}^3} + 3\left( {p + h} \right)} \right) - \left( {4{p^3} + 3p} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{4{p^3} + 12{p^2}h + 12p{h^2} + 4{h^3} + 3p + 3h - 4{p^3} - 3p}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{12{p^2}h + 12p{h^2} + 4{h^3} + 3h}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{5th + 2.5{h^2} + 6h}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {12{p^2} + 12ph + 4{h^2} + 3} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \left( {12{p^2} + 12ph + 4{h^2} + 3} \right)\\ & = 12{p^2} + 3\end{aligned}\)

The domain of the function \(f\) is \(\mathbb{R}\).

The domain of the derivative \(f'\) is \(\mathbb{R}\).

Thus, the domain of the function \(f\) and derivative \(f'\) is \(\mathbb{R}\).