Question

LET f (x) =4(3x^2 + 1)^2EVALUATE f'(1)

Answer 1

f'(1) = 192

Step-by-step explanation:

First, we need to solve the rigth side of f(x) as:

`\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}`

Now, we can derivate f(x) using the following:

`\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}`

Where a, b, and c are constants.

It means that f'(x) is equal to:

`\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}`

So, replacing x by 1, we get that f'(1) is equal to:

`\begin{gathered} f^(\prime)(1)=144\cdot1^3+48\cdot1 \\ f^(\prime)(1)=144+48 \\ f^(\prime)(1)=192 \end{gathered}`

Therefore, the answer is f'(1) = 192