Explanation

(c)
$f^{'}(x) = \frac{dx}{\left( 1 + x^{2} \right)^{\frac{3}{2}}}$
On integrating both sides,
$f(x) = \int_{}^{}\frac{dx}{\left( 1 + x^{2} \right)^{\frac{3}{2}}} + c$
Put $x = \tan{\theta \Rightarrow}dx = \sec^{2}\theta\ d\theta$
$\therefore\ f(x) = \int_{}^{}\frac{\sec^{2}\theta}{\sec^{3}\theta}\ d\theta + c = \int_{}^{}{\cos\theta}\ d\theta + c$
$\Rightarrow \ f(x) = \sin\theta + c$
$\Rightarrow f(x) = \frac{x}{\sqrt{1 + x^{2}}} + c$
$\Rightarrow f(0) = 0 + c \Rightarrow c = 0$
$\therefore f(x) = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow f(1) = \frac{1}{\sqrt{2}}$