Question

If $f(x) = \cot^{- 1}\left( \frac{x^{x} - x^{- x}}{2} \right),$ then $f'(1)$ is equal to

• A. – 1
• B. 1
• C. $\log{}2$
• D. $- \log 2$

Answer 1

Answer: (A)

– 1

Answer 2

$f(x) = \cot^{- 1}\left( \frac{x^{x} - x^{- x}}{2} \right)$

Put $x^{x} = \tan\theta$, ∴ $y = f(x) = \cot^{- 1}\left( \frac{\tan^{2}\theta - 1}{2\tan\theta} \right)$

= $\cot^{- 1}( - \cot 2\theta)$ = $\pi - \cot^{- 1}(\cot 2\theta)$

⇒ y = $\pi - 2\theta$ = $\pi - 2\tan^{- 1}(x^{x})$

⇒ $\frac{dy}{dx} = \frac{- 2}{1 + x^{2x}}.x^{x}(1 + \log x)$ ⇒ $f^{'}(1) = - 1$.