Let $$ f(x)=\sqrt{3 x^{2}+4 x+3} $$ $$ f^{\prime}(x)= $$ $$ f^{\prime}(2)= $$

Question

UpStudy AI · Accepted Answer

We are given the function f(x) = √(3x² + 4x + 3). To find f '(x), we'll first rewrite the function as:

f(x) = (3x² + 4x + 3)^(1/2).

Using the chain rule, the derivative is:

f '(x) = (1/2)(3x² + 4x + 3)^(-1/2) · (6x + 4)
       = (6x + 4) / [2√(3x² + 4x + 3)].

Now, to evaluate f '(2), substitute x = 2:

Numerator: 6(2) + 4 = 12 + 4 = 16.
  Denominator: 2√(3(2)² + 4(2) + 3) = 2√(12 + 8 + 3) = 2√23.

Thus, we have:

f '(2) = 16 / (2√23) = 8 / √23.

So the answers are:

f '(x) = (6x + 4) / [2√(3x² + 4x + 3)]
  f '(2) = 8 / √23.
$$ f'(x) = \frac{6x + 4}{2\sqrt{3x^2 + 4x + 3}} $$ $$ f'(2) = \frac{8}{\sqrt{23}} $$

Let \( f(x)=\sqrt{3 x^{2}+4 x+3} \) \( f^{\prime}(x)= \) \( f^{\prime}(2)= \)