Q88 / 170 · Manipal (MET) 2018 · Mathematics
If $f(x)=\frac{x+\cos x}{x-\cos x}, f^{\prime}\left(\frac{\pi}{2}\right)$ is equal to
- A$\frac{2(x \sin x+\cos x)}{(x-\cos x)^2}$
- B$\frac{2(x \cos x+\sin x)}{(x-\cos x)^2}$
- Correct answerC$\frac{-2(x \sin x+\cos x)}{(x+\cos x)^2}$
- DNone of these
Solution
Given, $f(x)=\frac{x+\cos x}{x-\cos x}$
On differentiating w.r.t. $x$, we get $\begin{aligned} f^{\prime}(x) & =\frac{\left[\begin{array}{c} (x-\cos x)(1-\sin x) \\ -(x+\cos x)(1+\sin x) \end{array}\right]}{(x-\cos x)^2} \\ & =\frac{-2 x \sin x-2 \cos x}{(x-\cos x)^2} \end{aligned}$
On differentiating w.r.t. $x$, we get $\begin{aligned} f^{\prime}(x) & =\frac{\left[\begin{array}{c} (x-\cos x)(1-\sin x) \\ -(x+\cos x)(1+\sin x) \end{array}\right]}{(x-\cos x)^2} \\ & =\frac{-2 x \sin x-2 \cos x}{(x-\cos x)^2} \end{aligned}$
Hence, the correct answer is option: 'C'.