`f(x)=((2^x+2^(-x))(tanx)sqrt(tan^-1(2x^2-3x+1)))/(7x^2-3x+1)^3`, then find f'(0)

Options

A. `sqrt(3pi)`
B. `sqrt(2pi)`
C. `sqrt(pi)`
D. `2sqrt(pi)`

Answer

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

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

we will use the definition of the derivative, which is given by:

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

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

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

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

Calculating each component:

- \( 2^0 = 1 \)
- \( 2^{-0} = 1 \)
- \( \tan(0) = 0 \)
- \( 2(0)^2 - 3(0) + 1 = 1 \) so \( \tan^{-1}(1) = \frac{\pi}{4} \) and \( \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2} \)
- \( 7(0)^2 - 3(0) + 1 = 1 \) so \( (1)^3 = 1 \)

Thus,

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

### Step 2: Calculate \( f(h) \)

Next, we compute \( f(h) \):

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

### Step 3: Substitute into the derivative formula

Now substituting \( f(h) \) 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 can simplify \( f(h) \):

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

### Step 5: Evaluate the limit

Now we need to evaluate the limit:

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

Using the properties of limits:

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

Thus, we have:

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

### Final Answer

Therefore,

\[ f'(0) = \sqrt{\pi} \]