If `f(x)=sin^(-1)""(2*(3)^(x))/(1+9^(x))`, then `f'(-(1)/(2))` is equal to

Options

A. `sqrt3log_(e)sqrt3`
B. `-sqrt3log_(e)sqrt3`
C. `-sqrt3log_(e)3`
D. `sqrt3log_(e)3`

Answer

To solve the problem, we need to find \( f'(-\frac{1}{2}) \) for the function \( f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right) \).

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right) \]
Notice that \( 9^x = (3^2)^x = (3^x)^2 \). Thus, we can rewrite the function as:
\[ f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + (3^x)^2}\right) \]

### Step 2: Use the identity for sine
We can use the identity for sine:
\[ \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)} \]
Let \( \theta = \tan^{-1}(3^x) \). Then:
\[ f(x) = \sin^{-1}(\sin(2\theta)) = 2\theta = 2\tan^{-1}(3^x) \]

### Step 3: Differentiate the function
Now we differentiate \( f(x) \):
\[ f'(x) = 2 \cdot \frac{d}{dx}(\tan^{-1}(3^x)) \]
Using the derivative of \( \tan^{-1}(u) \):
\[ \frac{d}{dx}(\tan^{-1}(u)) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = 3^x \) and \( \frac{du}{dx} = 3^x \ln(3) \). Therefore:
\[ f'(x) = 2 \cdot \frac{1}{1 + (3^x)^2} \cdot (3^x \ln(3)) \]
This simplifies to:
\[ f'(x) = \frac{2 \cdot 3^x \ln(3)}{1 + 9^x} \]

### Step 4: Evaluate at \( x = -\frac{1}{2} \)
Now we need to evaluate \( f'(-\frac{1}{2}) \):
\[ f'(-\frac{1}{2}) = \frac{2 \cdot 3^{-\frac{1}{2}} \ln(3)}{1 + 9^{-\frac{1}{2}}} \]
Calculating \( 3^{-\frac{1}{2}} = \frac{1}{\sqrt{3}} \) and \( 9^{-\frac{1}{2}} = \frac{1}{3} \):
\[ f'(-\frac{1}{2}) = \frac{2 \cdot \frac{1}{\sqrt{3}} \ln(3)}{1 + \frac{1}{3}} = \frac{2 \cdot \frac{1}{\sqrt{3}} \ln(3)}{\frac{4}{3}} = \frac{2 \cdot 3}{4 \sqrt{3}} \ln(3) = \frac{\sqrt{3}}{2} \ln(3) \]

### Final Result
Thus, the value of \( f'(-\frac{1}{2}) \) is:
\[ \frac{\sqrt{3}}{2} \ln(3) \]