`"Let "f(x)=((2^(x)+2^(-x))sin x sqrt(tan^(-1)(x^(2)-x+1)))/((7x^(2)+3x+1)^(3))`. Then find the value of f'(0).

Explanation

To find the derivative \( f'(0) \) for the function

\[ f(x) = \frac{(2^x + 2^{-x}) \sin x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3}, \]

we will use the definition of the derivative:

\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}. \]

### Step 1: Calculate \( f(0) \)

First, we need to evaluate \( f(0) \):

\[ f(0) = \frac{(2^0 + 2^{0}) \sin(0) \sqrt{\tan^{-1}(0^2 - 0 + 1)}}{(7(0)^2 + 3(0) + 1)^3}. \]

Calculating each term:

- \( 2^0 + 2^0 = 1 + 1 = 2 \)
- \( \sin(0) = 0 \)
- \( \tan^{-1}(0^2 - 0 + 1) = \tan^{-1}(1) = \frac{\pi}{4} \)
- Therefore, \( \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2} \)
- The denominator becomes \( (7(0)^2 + 3(0) + 1)^3 = 1^3 = 1 \)

Putting it all together:

\[ f(0) = \frac{2 \cdot 0 \cdot \frac{\sqrt{\pi}}{2}}{1} = 0. \]

### Step 2: Set Up the Limit

Now we set up the limit for the derivative:

\[ f'(0) = \lim_{x \to 0} \frac{f(x) - 0}{x} = \lim_{x \to 0} \frac{f(x)}{x}. \]

### Step 3: Substitute \( f(x) \)

Substituting \( f(x) \):

\[ f'(0) = \lim_{x \to 0} \frac{(2^x + 2^{-x}) \sin x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^3 \cdot x}. \]

### Step 4: Evaluate the Limit

Now, we evaluate the limit as \( x \to 0 \):

1. **Evaluate \( 2^x + 2^{-x} \)**:
\[ 2^x + 2^{-x} \to 1 + 1 = 2 \text{ as } x \to 0. \]

2. **Evaluate \( \sin x \)**:
\[ \sin x \to x \text{ as } x \to 0. \]

3. **Evaluate \( \tan^{-1}(x^2 - x + 1) \)**:
\[ x^2 - x + 1 \to 1 \text{ as } x \to 0 \implies \tan^{-1}(1) = \frac{\pi}{4}. \]
Thus, \( \sqrt{\tan^{-1}(x^2 - x + 1)} \to \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2} \).

4. **Evaluate the denominator**:
\[ (7x^2 + 3x + 1)^3 \to 1^3 = 1 \text{ as } x \to 0. \]

Putting it all together in the limit:

\[ f'(0) = \lim_{x \to 0} \frac{2 \cdot x \cdot \frac{\sqrt{\pi}}{2}}{1 \cdot x} = \lim_{x \to 0} \frac{2 \cdot x \cdot \frac{\sqrt{\pi}}{2}}{x} = \sqrt{\pi}. \]

### Final Answer

Thus, the value of \( f'(0) \) is

\[ \sqrt{\pi}. \]