Question

If $f(x) = \dfrac{x + \cos x}{x - \cos x}$, then $f'\!\left(\dfrac{\pi}{2}\right)$ equals

• A. $\dfrac{2}{\pi}$
• B. $\dfrac{-2}{\pi}$
• C. $\dfrac{4}{\pi^2}$
• D. $\dfrac{-4}{\pi^2}$

Hint:

Quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$. Evaluate trigonometric values carefully at the given point.

Answer 1

The Correct Option is D

Step 1: Conceptual Understanding:
Apply the quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
Step 2: Explanation in Detail:
$u = x+\cos x,\ u' = 1-\sin x$;\quad $v = x-\cos x,\ v' = 1+\sin x$.
At $x = \pi/2$: $\cos(\pi/2) = 0$, $\sin(\pi/2)=1$, so $u = \pi/2$, $u'=0$, $v=\pi/2$, $v'=2$.
$f'(\pi/2) = \dfrac{0\cdot(\pi/2) - (\pi/2)\cdot2}{(\pi/2)^2} = \dfrac{-\pi}{\pi^2/4} = -\dfrac{4}{\pi}$.
Step 3: Therefore, Stating the Final Answer
$f'(\pi/2) = -\dfrac{4}{\pi}$.