Respuesta :
Answer:
f'(1) = 192
Explanation:
First, we need to solve the rigth side of f(x) as:
[tex]\begin{gathered} f(x)=4(3x^2+1)^2 \\ f(x)=4((3x^2)^2+(2\cdot3x^2\cdot1)+1^2) \\ f(x)=4(9x^4+6x^2+1) \\ f(x)=36x^4+24x^2+4 \end{gathered}[/tex]Now, we can derivate f(x) using the following:
[tex]\begin{gathered} \text{if f(x)=b}\cdot\text{x}^a,\text{ then: f'(x) = b}\cdot a\cdot x^{a-1} \\ \text{If }f(x)=c,\text{ then f'(x) =0} \end{gathered}[/tex]Where a, b, and c are constants.
It means that f'(x) is equal to:
[tex]\begin{gathered} f^{\prime}(x)=36\cdot4\cdot x^{4-1}+24\cdot2\cdot x^{2-1}+0 \\ f^{\prime}(x)=144x^3+48x \end{gathered}[/tex]So, replacing x by 1, we get that f'(1) is equal to:
[tex]\begin{gathered} f^{\prime}(1)=144\cdot1^3+48\cdot1 \\ f^{\prime}(1)=144+48 \\ f^{\prime}(1)=192 \end{gathered}[/tex]Therefore, the answer is f'(1) = 192
