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 find \( f'(-\frac{1}{2}) \) for the function \( f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right) \), we will follow these steps:

### 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 denominator:
\[ f(x) = \sin^{-1}\left(\frac{2 \cdot 3^x}{1 + (3^x)^2}\right) \]

### Step 2: Differentiate using the chain rule
To differentiate \( f(x) \), we use the chain rule:
\[ f'(x) = \frac{1}{\sqrt{1 - \left(\frac{2 \cdot 3^x}{1 + 9^x}\right)^2}} \cdot \frac{d}{dx}\left(\frac{2 \cdot 3^x}{1 + 9^x}\right) \]

### Step 3: Differentiate the inner function
Let \( u = \frac{2 \cdot 3^x}{1 + 9^x} \). We need to find \( \frac{du}{dx} \):
Using the quotient rule:
\[ \frac{du}{dx} = \frac{(1 + 9^x)(2 \cdot 3^x \ln(3)) - 2 \cdot 3^x (9^x \ln(9))}{(1 + 9^x)^2} \]

### Step 4: Evaluate \( f'(-\frac{1}{2}) \)
Now we substitute \( x = -\frac{1}{2} \):
\[ 3^{-\frac{1}{2}} = \frac{1}{\sqrt{3}}, \quad 9^{-\frac{1}{2}} = \frac{1}{3} \]
Thus:
\[ u = \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{2/\sqrt{3}}{4/3} = \frac{2 \cdot 3}{4 \sqrt{3}} = \frac{3}{2 \sqrt{3}} \]

### Step 5: Substitute back into the derivative
Now we compute:
\[ f'(-\frac{1}{2}) = \frac{1}{\sqrt{1 - \left(\frac{3}{2 \sqrt{3}}\right)^2}} \cdot \frac{du}{dx} \text{ at } x = -\frac{1}{2} \]

### Step 6: Simplify the expression
Calculating \( \left(\frac{3}{2 \sqrt{3}}\right)^2 = \frac{9}{12} = \frac{3}{4} \):
\[ 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow \sqrt{1 - \left(\frac{3}{2 \sqrt{3}}\right)^2} = \frac{1}{2} \]

### Step 7: Final computation
Thus:
\[ f'(-\frac{1}{2}) = 2 \cdot \frac{du}{dx} \text{ at } x = -\frac{1}{2} \]
After evaluating \( \frac{du}{dx} \) at \( x = -\frac{1}{2} \) and simplifying, we find:
\[ f'(-\frac{1}{2}) = \frac{3}{2} \cdot \frac{1}{\sqrt{3}} \cdot \ln(3) = \frac{3 \ln(3)}{2 \sqrt{3}} \]

### Final Answer
Thus, the final answer is:
\[ f'(-\frac{1}{2}) = \sqrt{3} \ln(\sqrt{3}) \]