Answer is: option2

Solution:

We are asked to find \( f' \left(\frac{3\pi^2}{4}\right) \), where \( g^{-1}(x) = f(x) \). From the derivative rule for inverses, we know:

\[ f'(y) = \frac{1}{g'(x)} \quad \text{where} \quad y = g(x) \]

Step 1: Find \( g'(x) \)

For \( g(x) = \cos(x) + 3x^2 \), we differentiate:

\[ g'(x) = -\sin(x) + 6x \]

Step 2: Evaluate \( g' \) at \( x = \frac{\pi}{2} \)

At \( x = \frac{\pi}{2} \):

\[ g'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) + 6\left(\frac{\pi}{2}\right) = -1 + 3\pi \]

Step 3: Find \( f'\left(\frac{3\pi^2}{4}\right) \)

Since \( g\left(\frac{\pi}{2}\right) = \frac{3\pi^2}{4} \), we have:

\[ f'\left(\frac{3\pi^2}{4}\right) = \frac{1}{g'\left(\frac{\pi}{2}\right)} = \frac{1}{-1 + 3\pi} \]

Final Answer:

The correct answer is: b) \( \frac{1}{3\pi - 1} \)