Suppose `f(x)=((2^(x)+2^(-x))tan x sqrt(tan^(-1)(x^(2)-x+1)))/((7x^(2)+3x+1)^(3))` .Then the value of `f'(0)` is equal to

Options

A. `sqrtx`
B. 0
C. `pi/2`
D. `sqrtpi`

Answer

To find the value of \( f'(0) \) for the function

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

we will use the definition of the derivative:

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

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

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

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

Calculating each part:

- \( 2^0 + 2^0 = 1 + 1 = 2 \)
- \( \tan(0) = 0 \)
- \( \tan^{-1}(0^2 - 0 + 1) = \tan^{-1}(1) = \frac{\pi}{4} \)
- 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: Calculate \( f(h) \)

Next, we need to find \( f(h) \):

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

### Step 3: Substitute into the derivative formula

Now we substitute \( f(h) \) and \( f(0) \) into the derivative formula:

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

### Step 4: Simplify \( f(h) \)

We need to analyze \( f(h) \) as \( h \to 0 \):

- As \( h \to 0 \), \( 2^h + 2^{-h} \to 2 \)
- \( \tan(h) \approx h \)
- \( h^2 - h + 1 \to 1 \) so \( \tan^{-1}(h^2 - h + 1) \to \tan^{-1}(1) = \frac{\pi}{4} \) and thus \( \sqrt{\tan^{-1}(h^2 - h + 1)} \to \frac{\sqrt{\pi}}{2} \)
- The denominator \( 7h^2 + 3h + 1 \to 1 \) so \( (7h^2 + 3h + 1)^3 \to 1^3 = 1 \)

Putting this together:

\[ f(h) \approx \frac{(2)(h)\left(\frac{\sqrt{\pi}}{2}\right)}{1} = h\sqrt{\pi} \]

### Step 5: Substitute back into the limit

Now we substitute back into our limit:

\[ f'(0) = \lim_{h \to 0} \frac{h\sqrt{\pi}}{h} = \sqrt{\pi} \]

### Conclusion

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

\[ \boxed{\sqrt{\pi}} \]