If `f(x)=sin^(-1)(3x-4x^(3)).` Then answer the following
The value of f'(0), is

Options

A. -3
B. 3
C. `sqrt2`
D. `-sqrt2`

Answer

To find the value of \( f'(0) \) for the function \( f(x) = \sin^{-1}(3x - 4x^3) \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx} \left( \sin^{-1}(3x - 4x^3) \right) \]
Using the derivative formula for \( \sin^{-1}(u) \), which is \( \frac{1}{\sqrt{1 - u^2}} \), where \( u = 3x - 4x^3 \):
\[ f'(x) = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot \frac{d}{dx}(3x - 4x^3) \]

### Step 2: Differentiate the inner function
Now we need to differentiate \( 3x - 4x^3 \):
\[ \frac{d}{dx}(3x - 4x^3) = 3 - 12x^2 \]

### Step 3: Substitute back into the derivative
Substituting this back into our expression for \( f'(x) \):
\[ f'(x) = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2) \]

### Step 4: Evaluate at \( x = 0 \)
Now we need to find \( f'(0) \):
\[ f'(0) = \frac{1}{\sqrt{1 - (3(0) - 4(0)^3)^2}} \cdot (3 - 12(0)^2) \]
This simplifies to:
\[ f'(0) = \frac{1}{\sqrt{1 - 0^2}} \cdot 3 = \frac{1}{\sqrt{1}} \cdot 3 = 3 \]

### Final Answer
Thus, the value of \( f'(0) \) is:
\[ \boxed{3} \]