If $y = an^{-1}\left[ \dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} ight]$ $(0

Question

If $y = 	an^{-1}\left[ \dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} ight]$ $(0 < |x| < 1)$, find $\dfrac{dy}{dx}$.

Accepted Answer

>

> [!TLDR]
> Substituting $x^2=\cos2\varphi$ simplifies the expression to $y=\frac{\pi}{4}+\frac12\cos^{-1}(x^2)$, giving $\dfrac{dy}{dx}=\frac{-x}{\sqrt{1-x^4}}$.
>
> **Step 1: Substitute $x^2=\cos2\varphi$**, i.e. $\varphi=\dfrac12\cos^{-1}(x^2)$
>
> Then $1+x^2 = 1+\cos2\varphi = 2\cos^2\varphi$ and $1-x^2 = 1-\cos2\varphi = 2\sin^2\varphi$, so
>
> $\sqrt{1+x^2}=\sqrt2\cos\varphi$, \quad $\sqrt{1-x^2}=\sqrt2\sin\varphi$
>
> **Step 2: Simplify the fraction inside $	an^{-1}$**
>
> $\dfrac{\sqrt2\cos\varphi+\sqrt2\sin\varphi}{\sqrt2\cos\varphi-\sqrt2\sin\varphi} = \dfrac{\cos\varphi+\sin\varphi}{\cos\varphi-\sin\varphi} = \dfrac{1+	an\varphi}{1-	an\varphi} = 	an\!\left(\dfrac{\pi}{4}+\varphiight)$
>
> **Step 3: Simplify y**
> …

Q.If $y = \tan^{-1}\left[ \dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right]$ $(0 < |x| < 1)$ , find $\dfrac{dy}{dx}$ .

Concept understanding — Rules of Differentiation

Q.If y=tan⁡−1[1+x2+1−x21+x2−1−x2]y = \tan^{-1}\left[ \dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right]y=tan−1[1+x2​−1−x2​1+x2​+1−x2​​] (0<∣x∣<1)(0 < |x| < 1)(0<∣x∣<1), find dydx\dfrac{dy}{dx}dxdy​.

🎓Unlock everything free for 14 days

Q.If $y = \tan^{-1}\left[ \dfrac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right]$ $(0 < |x| < 1)$ , find $\dfrac{dy}{dx}$ .

Unlock everything free for 14 days