Question

Let $f\left(x\right)=2{\tan }^{-1}x+{\sin }^{-1}\frac{2x}{1+x^2}$. Then

• A. $f^{{}^{\prime }}\left(2\right)=f^{{}^{\prime }}\left(3\right)$
• B. $f^{{}^{\prime }}\left(2\right)=0$
• C. $f^{{}^{\prime }}\left(\frac{1}{2}\right)=\frac{16}{5}$
• D. $f^{{}^{\prime }}\left(\frac{1}{2}\right)=0$

Answer 1

Answer: (C)

$f^{{}^{\prime }}\left(\frac{1}{2}\right)=\frac{16}{5}$

$f\left(x\right)=2{\tan }^{-1}x+{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$
$=2\frac{1}{x^2+1}+\frac{\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)}{\sqrt{1-\frac{{4x}^2}{(x^2+1^2)}}}$
$=\frac{2}{x^2+1}+\frac{2(1+x^2)}{(1-x^2)\sqrt{1-\frac{4x^2}{{(1+x^2)}^2}}}$
$f^{\prime }\left(\frac{1}{2}\right)=\frac{2}{{\left(\frac{1}{2}\right)}^2+1}+\frac{2(1-{\left(\frac{1}{2}\right)}^2)}{{(1+{\frac{1}{2}}^2)}^2\sqrt{1-\frac{4{\left(\frac{1}{2}\right)}^2}{{(1+{\left(\frac{1}{2}\right)}^2)}^2}}}=\frac{16}{5}$