Question

$y=tan^{-1}(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}})$

Answer 1

$\frac{dy}{dx}=0$.

Answer 2

Solution:

We are given

$$y=\tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right).$$

Note that for all $x$ (except at points where the expression is undefined),

$$\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}=1.$$

Thus,

$$y=\tan^{-1}(1)=\frac{\pi}{4},$$

which is a constant. Hence,

$$\frac{dy}{dx}=0.$$

---

Explanation (Minimal):

Since the numerator and denominator are identical (except where undefined), the inner expression simplifies to 1. So, $y=\tan^{-1}(1)=\frac{\pi}{4}$ (a constant), and its derivative is 0.