Question

The slope of the tangent to the curve: $\tan y=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$at $x=\frac{1}{2}$

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

$\frac{d}{dx}\left(\arctan \left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\right)$ $=\frac{d}{du}\left(\arctan \left(u\right)\right)\frac{d}{dx}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

$\frac{d}{du}\left(\arctan \left(u\right)\right)$ $=\frac{1}{u^2+1}$

$=\frac{1}{u^2+1}\cdot \frac{2}{{(\sqrt{1+x}+\sqrt{1-x})}^2\sqrt{x+1}\sqrt{-x+1}}$

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

$=\frac{1}{{\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)}^2+1}\cdot \frac{2}{{(\sqrt{1+x}+\sqrt{1-x})}^2\sqrt{x+1}\sqrt{-x+1}}$

$=\frac{1}{2\sqrt{x+1}\sqrt{-x+1}}$

$f^{\prime }\left(\frac{1}{2}\right)=\frac{1}{2\sqrt{\frac{1}{2}+1}\sqrt{-\frac{1}{2}+1}}=\frac{1}{\sqrt{3}}$