Correct Answer: Option A

Explanation

Let \( u = 3x^2 \) and \( v = (x^3 + 1)^{1/2} \)

Then \( y = u \cdot v \)

- \( \frac{du}{dx} = 6x \)

- For \( v \):  
  \(\frac{dv}{dx} = \frac{1}{2}(x^3 + 1)^{-1/2} \cdot 3x^2 = \frac{3x^2}{2\sqrt{x^3 + 1}}\)

Now apply the product rule:  
\(\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \)

\(\frac{dy}{dx} = 3x^2 \cdot \frac{3x^2}{2\sqrt{x^3 + 1}} + \sqrt{x^3 + 1} \cdot 6x\)

\(= \frac{9x^4}{2\sqrt{x^3 + 1}} + 6x \sqrt{x^3 + 1}\)

Simplify (common denominator \( 2\sqrt{x^3 + 1} \)):

\(\frac{dy}{dx} = \frac{9x^4 + 12x (x^3 + 1)}{2\sqrt{x^3 + 1}}\)

\(= \frac{9x^4 + 12x^4 + 12x}{2\sqrt{x^3 + 1}} = \frac{21x^4 + 12x}{2\sqrt{x^3 + 1}}\)

Factor out 3x:

\(\frac{dy}{dx} = \frac{3x(7x^3 + 4)}{2\sqrt{x^3 + 1}}\)

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