Understanding the Derivative Problem

We are asked to find the derivative of the function \({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}}} \right)\) with respect to \({\rm{x}}\). This involves differentiating an inverse trigonometric function. The structure of the expression inside the inverse cosine suggests using a trigonometric substitution to simplify the function before differentiation.

Applying Trigonometric Substitution

Let's observe the term \( \frac{{2{\rm{x}}}}{{1 + {{\rm{x}}^2}}} \). This form is related to the double angle formula for sine, \( \sin(2\theta) = \frac{2 \tan \theta}{1 + \tan^2 \theta} \). This suggests substituting \( {\rm{x}} = \tan \theta \).

If \( {\rm{x}} = \tan \theta \), then \( \theta = {\tan ^{ - 1}}{\rm{x}} \). We usually consider the principal value branch for \( {\tan ^{ - 1}}{\rm{x}} \), which means \( - \frac{\pi }{2} < \theta < \frac{\pi }{2} \).

Simplifying the Function Using Substitution

Substitute \( {\rm{x}} = \tan \theta \) into the given function:

\({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)\)

Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \) and \( \frac{\tan \theta}{\sec^2 \theta} = \frac{\sin \theta / \cos \theta}{1/\cos^2 \theta} = \sin \theta \cos \theta \), we get:

\({\rm{y}} = {\cos ^{ - 1}}\left( {\frac{{2\tan \theta }}{{{{\sec }^2}\theta }}} \right) = {\cos ^{ - 1}}\left( {2\sin \theta \cos \theta } \right)\)

Now, using the double angle identity \( 2\sin \theta \cos \theta = \sin(2\theta) \):

\({\rm{y}} = {\cos ^{ - 1}}\left( {\sin(2\theta) } \right)\)

Transforming to Cosine

To simplify \( {\cos ^{ - 1}}\left( {\sin(2\theta) } \right) \), we can convert \( \sin(2\theta) \) to a cosine function using the identity \( \sin A = \cos\left(\frac{\pi}{2} - A\right) \).

So, \( \sin(2\theta) = \cos\left(\frac{\pi}{2} - 2\theta\right) \).

The function becomes \( {\rm{y}} = {\cos ^{ - 1}}\left( {\cos\left(\frac{\pi}{2} - 2\theta\right) } \right) \).

Evaluating the Inverse Cosine for a Specific Range

The expression \( {\cos ^{ - 1}}\left( {\cos u} \right) \) simplifies to \( u \) only when \( 0 \le u \le \pi \). In our case, \( u = \frac{\pi}{2} - 2\theta \). We need to consider the range of \( u \) based on the given condition on \({\rm{x}}\).

The question provides an option that specifies the condition \( |{\rm{x}}| < 1 \). Let's analyze this condition.

\( |{\rm{x}}| < 1 \implies -1 < {\rm{x}} < 1 \).

Since \( {\rm{x}} = \tan \theta \) and \( - \frac{\pi }{2} < \theta < \frac{\pi }{2} \), the condition \( -1 < \tan \theta < 1 \) implies \( - \frac{\pi }{4} < \theta < \frac{\pi }{4} \).

Now let's find the range of \( 2\theta \):

\( - \frac{\pi }{4} < \theta < \frac{\pi }{4} \implies - \frac{\pi }{2} < 2\theta < \frac{\pi }{2} \).

Next, let's find the range of \( u = \frac{\pi}{2} - 2\theta \):

\( - \frac{\pi }{2} < 2\theta < \frac{\pi }{2} \implies - \frac{\pi }{2} < -2\theta < \frac{\pi }{2} \) (multiplying by -1 reverses the inequalities)

\( \frac{\pi}{2} - \frac{\pi}{2} < \frac{\pi}{2} - 2\theta < \frac{\pi}{2} + \frac{\pi}{2} \) (adding \( \frac{\pi}{2} \))

\( 0 < \frac{\pi}{2} - 2\theta < \pi \)

So, for \( |{\rm{x}}| < 1 \), the argument \( u = \frac{\pi}{2} - 2\theta \) is in the range \( (0, \pi) \), which is within the principal range \( [0, \pi] \) for \( {\cos ^{ - 1}} \). Therefore, under this condition, \( {\cos ^{ - 1}}\left( {\cos u} \right) = u \).

\({\rm{y}} = {\cos ^{ - 1}}\left( {\cos\left(\frac{\pi}{2} - 2\theta\right) } \right) = \frac{\pi}{2} - 2\theta \)

Substituting Back and Differentiating

Now substitute back \( \theta = {\tan ^{ - 1}}{\rm{x}} \):

\({\rm{y}} = \frac{\pi}{2} - 2{\tan ^{ - 1}}{\rm{x}} \)

Now differentiate \( {\rm{y}} \) with respect to \( {\rm{x}} \):

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{d}{{d{\rm{x}}}}\left( {\frac{\pi}{2} - 2{\tan ^{ - 1}}{\rm{x}}} \right) \)

Using the linearity of differentiation and the standard derivative formula \( \frac{d}{{d{\rm{x}}}}({\tan ^{ - 1}}{\rm{x}}) = \frac{1}{{1 + {{\rm{x}}^2}}} \), we get:

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{d}{{d{\rm{x}}}}\left( {\frac{\pi}{2}} \right) - 2\frac{d}{{d{\rm{x}}}}\left( {{\tan ^{ - 1}}{\rm{x}}} \right) \)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 0 - 2\left( {\frac{1}{{1 + {{\rm{x}}^2}}}} \right) \)

\(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = - \frac{2}{{1 + {{\rm{x}}^2}}} \)

This result is valid for the condition \( |{\rm{x}}| < 1 \).

Conclusion

Based on the derivation using trigonometric substitution and considering the condition \( |{\rm{x}}| < 1 \), the derivative \( \frac{{{\rm{dy}}}}{{{\rm{dx}}}} \) is \( - \frac{2}{{1 + {{\rm{x}}^2}}} \).

Revision Table: Key Concepts in Differentiation

Additional Information on Inverse Trigonometric Function Derivatives

Inverse trigonometric functions, also known as arcus functions, are the inverse functions of the trigonometric functions. Their derivatives are essential in calculus, especially for integration techniques and solving differential equations.

Key inverse trigonometric functions and their derivatives:

• \( \frac{d}{{d{\rm{x}}}}({\sin ^{ - 1}}{\rm{x}}) = \frac{1}{{\sqrt{1 - {{\rm{x}}^2}}}} \), for \( -1 < {\rm{x}} < 1 \)
• \( \frac{d}{{d{\rm{x}}}}({\cos ^{ - 1}}{\rm{x}}) = - \frac{1}{{\sqrt{1 - {{\rm{x}}^2}}}} \), for \( -1 < {\rm{x}} < 1 \)
• \( \frac{d}{{d{\rm{x}}}}({\tan ^{ - 1}}{\rm{x}}) = \frac{1}{{1 + {{\rm{x}}^2}}} \), for all real \( {\rm{x}} \)
• \( \frac{d}{{d{\rm{x}}}}({\cot ^{ - 1}}{\rm{x}}) = - \frac{1}{{1 + {{\rm{x}}^2}}} \), for all real \( {\rm{x}} \)
• \( \frac{d}{{d{\rm{x}}}}({\sec ^{ - 1}}{\rm{x}}) = \frac{1}{|{\rm{x}}|\sqrt{{{\rm{x}}^2} - 1}} \), for \( |{\rm{x}}| > 1 \)
• \( \frac{d}{{d{\rm{x}}}}({\csc ^{ - 1}}{\rm{x}}) = - \frac{1}{|{\rm{x}}|\sqrt{{{\rm{x}}^2} - 1}} \), for \( |{\rm{x}}| > 1 \)

It's crucial to remember that the derivatives of inverse trigonometric functions often have domain restrictions, which arise from the domain of the original trigonometric function's restricted branch used to define the inverse.