Question

If $y={\tan }^{-1}\left[\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}-\sqrt{1-x^3}}\right]$, then $y^{\prime }$ at $x=0$ is

• A. $0$
• B. $-\frac{3}{2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is A $0$

$\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}-\sqrt{1-x^3}}=\left(\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}-\sqrt{1-x^3}}\right)\left(\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}+\sqrt{1-x^3}}\right)=\frac{{(\sqrt{1+x^3}+\sqrt{1-x^3})}^2}{{\left(\sqrt{1+x^3}\right)}^2-{\left(\sqrt{1-x^3}\right)}^2}=\frac{1+x^3+1-x^3+2\sqrt{(1+x^3)(1-x^3)}}{(1+x^3)-(1-x^3)}=\frac{2-2x^3+2\sqrt{1-x^6}}{2x^3}=\frac{1-x^3+\sqrt{1-x^6}}{x^3}\Rightarrow y={\tan }^{-1}\left[\frac{1-x^3+\sqrt{1-x^6}}{x^3}\right]\frac{dy}{dx}=\frac{1}{1+\frac{{(1-x^3+\sqrt{1-x^6})}^2}{x^6}}\times \frac{-3x^2(1+\frac{1}{\sqrt{1+x^6}})}{x^6}\Rightarrow {\left|\frac{dy}{dx}\right|}_{atx=0}=0$