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

Given the function \( f(x) = \frac{(2^x + 2^{-x}) \tan x \sqrt{\tan^{-1}(x^2 - x + 1)}}{(7x^2 + 3x + 1)^{3}} \), we aim to determine the value of \( f'(0) \).

This requires applying differentiation rules to find \( f'(x) \) and subsequently evaluating it at \( x = 0 \). Simplification of the expression prior to differentiation is advisable.

Step 1: Analyze the function's components at \( x = 0 \).

• At \( x = 0 \), \( 2^x + 2^{-x} \) becomes \( 2^0 + 2^0 = 1 + 1 = 2 \).
• At \( x = 0 \), \( \tan x \) is \( \tan 0 = 0 \).
• At \( x = 0 \), \( \sqrt{\tan^{-1}(x^2 - x + 1)} \) is \( \sqrt{\tan^{-1}(1)} = \sqrt{\frac{\pi}{4}} = \frac{\sqrt{\pi}}{2} \).
• At \( x = 0 \), \( (7x^2 + 3x + 1)^3 \) is \( 1^3 = 1 \).

Substituting these values into the function at \( x = 0 \):

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

This confirms that \( f(0) = 0 \).

Step 2: Differentiate the function using quotient rule.

The function is a quotient. Applying the quotient rule \(\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\):

Let:

• \( u = (2^x + 2^{-x}) \tan x \sqrt{\tan^{-1}(x^2 - x + 1)} \)
• \( v = (7x^2 + 3x + 1)^3 \)

Differentiating \( u(x) \) and \( v(x) \) at \( x=0 \):

• As components of \( u(x) \) and \( v(x) \) simplify to constants or zero at \( x=0 \), differentiation proceeds linearly.
• The term \( \sqrt{\tan^{-1}(x^2 - x + 1)} \) contributes to the derivative's non-zero value near \( x=0 \).

\( f'(0) \) is determined by the higher-order derivatives of the components. The term \( \frac{\sqrt{\pi}}{2} \) plays a crucial role. After simplification and elimination of terms that become zero at \( x=0 \), the derivative's form is dominated by this component.

Conclusion: The derivative calculation indicates a result proportional to \(\sqrt{\pi}\).

Thus, the answer is \(\sqrt\pi\).