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 $x=\frac{1}{2}$ is?

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

Answer 1

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

$\tan y=\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}$
Divide numerator and denominator by $\sqrt{1+x}$ on RHS
$\tan y=\frac{1-\sqrt{\frac{1-x}{1+x}}}{1+\sqrt{\frac{1-x}{1+x}}}$
$y={\tan }^{-1}\left[\frac{1-\sqrt{\frac{1-x}{1+x}}}{1+\sqrt{\frac{1-x}{1+x}}}\right]$
$y={\tan }^{-1}\left(1\right)-{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$
$y=\frac{\pi }{4}-{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$
diff. on both sides
$\frac{dy}{dx}=0-\frac{1}{1+\left(\frac{1-x}{1+x}\right)}\left[\frac{-\frac{\sqrt{1+x}}{2\sqrt{1-x}}-\frac{\sqrt{1-x}}{2\sqrt{1+x}}}{(1+x)}\right]$
$=\left[\frac{1+x}{2}\right]\left[\frac{1+x+1-x}{1(1+x)\sqrt{1-x^2}}\right]=\left[\frac{1+x}{2}\right]\frac{2}{2(1+x)\sqrt{1-x^2}}$
$=\frac{1}{2\sqrt{1-x^2}}$
${\left(\frac{dy}{dx}\right)}_{x=\frac{1}{2}}=\frac{1}{2\sqrt{1-\frac{1}{4}}}=\frac{2}{2\sqrt{3}}=\frac{1}{\sqrt{3}}$