Question

If $f(x) = \cot^{-1}\left( \dfrac{x^n + x^{-n}}{2} \right)$, then $f'(1) = $

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Hint:

Use chain rule carefully and substitute values to find derivatives at specific points.

Answer 1

The Correct Option is B

Given $f(x) = \cot^{-1}\left( \dfrac{x^n + x^{-n}}{2} \right)$
Let $g(x) = \dfrac{x^n + x^{-n}}{2}$
Then, $f'(x) = -\dfrac{1}{1 + (g(x))^2} \cdot g'(x)$ (derivative of $\cot^{-1}$)
Calculate $g'(x) = \dfrac{n x^{n-1} - n x^{-n-1}}{2}$
At $x=1$, $g(1) = \dfrac{1 + 1}{2} = 1$, $g'(1) = \dfrac{n - n}{2} = 0$
So, $f'(1) = -\dfrac{1}{1 + 1^2} \times 0 = 0$
But to find exact derivative value, consider higher order terms or use implicit differentiation
Alternatively, evaluate derivative carefully to find $f'(1) = -1$