Question

Suppose\(f(x)=\frac{(2^x+2^{-x})tanx\sqrt{tan^{-1}(x^2-x+1)}}{(7x^2+3x+1)^{3}}\) then the value of \(f'(0)\) is equal to

• A. \(\pi\)
• B. 0
• C. \(\sqrt\pi\)
• D. \(\frac{\pi}{2}\)

Answer 1

The Correct Option is C

Step 1: Calculate \(f'(0)\) Using the Definition of Derivative

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

Step 2: Evaluate \(f(h) - f(0)\)

Substitute \(x = h\) and \(x = 0\) into \(f(x)\):

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

Step 3: Simplify the Expression

Using the limit properties, we get:

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

So, the correct answer is: \(\sqrt{\pi}\)