If $y = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {\sec ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$, then $\dfrac{{dy}}{{dx}} = $
(A) $\dfrac{4}{{\left( {1 - {x^2}} \right)}}$
(B) $\dfrac{4}{{\left( {1 + {x^2}} \right)}}$
(C) $\dfrac{1}{{\left( {1 + {x^2}} \right)}}$
(D) $\dfrac{{ - 4}}{{\left( {1 + {x^2}} \right)}}$

Hint: In the given problem, we are required to differentiate the function provided to us in the question with respect to x. Since, $y = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {\sec ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$ is a complex composite function, so we will have to apply chain rule of differentiation in the process of differentiating $y = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {\sec ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$. So, differentiation with respect to x will be done layer by layer using the chain rule of differentiation. We will first simplify the given function by substituting x as a tangent of an angle. Then, we will express use trigonometric formulae such as $\sin 2\theta = \left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$. Also the derivative of ${\tan ^{ - 1}}x$ with respect to $x$ must be remembered.

Complete answer:
So, we have, $y = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {\sec ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$
Now, we assume $x = \tan \theta $ in the function to simplify the expression.
So, we get, \[y = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + {\sec ^{ - 1}}\left( {\dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}} \right)\]
Now, we know the double angle trigonometric identities of sine and cosine as \[\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }} = \sin 2\theta \] and \[\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} = \cos 2\theta \]. So, using these formulae in the function, we get,
\[ \Rightarrow y = {\sin ^{ - 1}}\left( {\sin \left( {2\theta } \right)} \right) + {\sec ^{ - 1}}\left( {\dfrac{1}{{\cos \left( {2\theta } \right)}}} \right)\]
Now, we know that cosine and secant functions are reciprocal trigonometric functions. So, we get,
\[ \Rightarrow y = {\sin ^{ - 1}}\left( {\sin \left( {2\theta } \right)} \right) + {\sec ^{ - 1}}\left( {\sec \left( {2\theta } \right)} \right)\]
We also know that ${\sin ^{ - 1}}\left( {\sin x} \right) = x$ and ${\sec ^{ - 1}}\left( {\sec x} \right) = x$ for x belonging to the principal value branch.
\[ \Rightarrow y = \left( {2\theta } \right) + \left( {2\theta } \right)\]
Adding up the like terms, we get,
\[ \Rightarrow y = 4\theta \]
Now, substituting back the value of $\theta $ as ${\tan ^{ - 1}}x$ since we assumed $x = \tan \theta $. So, we get,
\[ \Rightarrow y = 4{\tan ^{ - 1}}\left( x \right)\]
Now, we differentiate both sides of the above function.
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {4{{\tan }^{ - 1}}\left( x \right)} \right]\]
We know that the derivative of the inverse tangent function ${\tan ^{ - 1}}x$ is $\dfrac{1}{{1 + {x^2}}}$ . So, we get the derivative as,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{1 + {x^2}}}\]
So, the value of \[\dfrac{{dy}}{{dx}}\] if $y = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {\sec ^{ - 1}}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$ is \[\left( {\dfrac{4}{{1 + {x^2}}}} \right)\].
Hence, option (B) is the correct answer.

Note:
The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The questions involve simplification of the function using trigonometric formulae and identity and then using basic results of differentiation to find the correct answer. We must take care while substituting the value of a variable.