ANSWER:
f'(1) = 7.5
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]f\left(x\right)=3x\sqrt{x^3}[/tex]
The first thing is to derive the function, just like this:
[tex]\begin{gathered} f^{\prime}(x)=3\frac{d}{dx}\left(x\sqrt{x^3}\right) \\ \\ f^{\prime}(x)=3\left(\frac{d}{dx}\left(x\right)\sqrt{x^3}+\frac{d}{dx}\left(\sqrt{x^3}\right)x\right) \\ \\ f^{\prime}(x)=3\left(1\cdot\sqrt{x^3}+\frac{3\sqrt{x}}{2}\cdot x\right) \\ \\ f^{\prime}(x)=3\sqrt{x^3}+\frac{9}{2}\sqrt{x^3} \\ \\ f^{\prime}(x)=\frac{15}{2}\sqrt{x^3} \end{gathered}[/tex]
Now, we evaluate when x = 1:
[tex]\begin{gathered} f^{\prime}(x)=\frac{15}{2}\sqrt{(1)^3} \\ \\ \begin{equation*} f^{\prime}(x)=\frac{15}{2} \end{equation*} \end{gathered}[/tex]
The result is equal to 15/2 or 7.5
