Question

If $y={\sin }^{-1}\frac{1}{\sqrt{1+x^2}}+{\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ then $\frac{dy}{dx}$ equals $(x\ne 0)$

• A. $\frac{1}{2(1+x^2)}$
• B. $-\frac{1}{2(1+x^2)}$
• C. $0$
• D. None of these

Answer 1

The correct option is B $-\frac{1}{2(1+x^2)}$

$\frac{d}{dx}\left(\arcsin \left(\frac{1}{\sqrt{1+x^2}}\right)\right)+\frac{d}{dx}\left(\arctan \left(\frac{\sqrt{1+x^2}-1}{x}\right)\right)$

$\frac{d}{dx}\left(\arcsin \left(\frac{1}{\sqrt{1+x^2}}\right)\right)$

$\frac{d}{du}\left(\arcsin \left(u\right)\right)\frac{d}{dx}\left(\frac{1}{\sqrt{1+x^2}}\right)$

$\frac{d}{du}\left(\arcsin \left(u\right)\right)$ $=\frac{1}{\sqrt{1-u^2}}$

$\frac{d}{dx}\left(\frac{1}{\sqrt{1+x^2}}\right)$

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

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

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

$\frac{d}{dx}(1+x^2)$ $=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x^2\right)$

$\frac{d}{dx}\left(1\right)$ $=0$ , $\frac{d}{dx}\left(x^2\right)$ $=2x^{2-1}$ $=2x$

$=(-\frac{1}{2u^{\frac{3}{2}}})\cdot 2x$

$Substitutebacku=1+x^2$

$=(-\frac{1}{2{(1+x^2)}^{\frac{3}{2}}})\cdot 2x$

$=-\frac{1\cdot 2x}{2{(1+x^2)}^{\frac{3}{2}}}$

$=-\frac{x}{{(1+x^2)}^{\frac{3}{2}}}$

$=\frac{1}{\sqrt{1-u^2}}(-\frac{x}{{(1+x^2)}^{\frac{3}{2}}})$

$Substitutebacku=\frac{1}{\sqrt{1+x^2}}$

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

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

$=-\frac{x}{{(x^2+1)}^{\frac{3}{2}}}\sqrt{\frac{x^2+1}{x^2}}$

$=-\frac{\frac{\sqrt{x^2+1}}{\sqrt{x^2}}x}{{(x^2+1)}^{\frac{3}{2}}}$

$=-\frac{x\sqrt{x^2+1}}{\sqrt{x^2}{(x^2+1)}^{\frac{3}{2}}}$

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

$\frac{d}{dx}\left(\arctan \left(\frac{\sqrt{1+x^2}-1}{x}\right)\right)$

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

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

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

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

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

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

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

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