Finding the Derivative of an Inverse Tangent Function

The question asks for the derivative of the function \(u = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1{\rm{\;}} + {{\rm{x}}^2}} - {\rm{\;}}1}}{{\rm{x}}}} \right)\) with respect to another function, \(v = {\tan ^{ - 1}}{\rm{x}}\). To find the derivative of \(u\) with respect to \(v\), we use the formula:

\[ \frac{{du}}{{dv}} = \frac{{du/dx}}{{dv/dx}} \]

First, let's simplify the function \(u(x)\) using a substitution. This makes finding \(du/dx\) much easier.

Simplifying u(x) Using Substitution

Let's set \(x = \tan \theta\). For the principal value branch of \({\tan ^{ - 1}}x\), we usually consider \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\). Substituting \(x = \tan \theta\) into the expression for \(u\):

\[ u = {\tan ^{ - 1}}\left( {\frac{{\sqrt {1{\rm{\;}} + {{\left( {\tan \theta} \right)}^2}} - {\rm{\;}}1}}{{\tan \theta}}} \right) \]

Using the trigonometric identity \(1 + {\tan ^2}\theta = {\sec ^2}\theta\):

\[ u = {\tan ^{ - 1}}\left( {\frac{{\sqrt {{\sec }^2}\theta} - {\rm{\;}}1}}{{\tan \theta}}} \right) \]

Since \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), \(\sec \theta = \frac{1}{{\cos \theta}}\) is positive, so \(\sqrt {{\sec }^2}\theta = |\sec\theta| = \sec\theta\).

\[ u = {\tan ^{ - 1}}\left( {\frac{{\sec\theta - {\rm{\;}}1}}{{\tan \theta}}} \right) \]

Now, express \(\sec\theta\) and \(\tan\theta\) in terms of \(\sin\theta\) and \(\cos\theta\):

\[ u = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\cos \theta}} - {\rm{\;}}1}}{{\frac{{\sin \theta}}{{\cos \theta}}}}} \right) \]

Combine the terms in the numerator:

\[ u = {\tan ^{ - 1}}\left( {\frac{{\frac{{1 - \cos \theta}}{{\cos \theta}}}}{{\frac{{\sin \theta}}{{\cos \theta}}}}} \right) \]

Cancel out the \(\cos \theta\) in the numerator and denominator:

\[ u = {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta}}{{\sin \theta}}} \right) \]

Using the half-angle identities, \(1 - \cos \theta = 2{\sin ^2}\left( {\frac{\theta}{2}} \right)\) and \(\sin \theta = 2\sin\left( {\frac{\theta}{2}} \right)\cos\left( {\frac{\theta}{2}} \right)\):

\[ u = {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\left( {\frac{\theta}{2}} \right)}}{{2\sin\left( {\frac{\theta}{2}} \right)\cos\left( {\frac{\theta}{2}} \right)}}} \right) \]

Simplify the expression:

\[ u = {\tan ^{ - 1}}\left( {\frac{{\sin\left( {\frac{\theta}{2}} \right)}}{{\cos\left( {\frac{\theta}{2}} \right)}}} \right) \] \[ u = {\tan ^{ - 1}}\left( {\tan\left( {\frac{\theta}{2}} \right)} \right) \]

Since \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\), it follows that \(-\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{4}\). In this range, \({\tan ^{ - 1}}(\tan y) = y\).

\[ u = \frac{\theta}{2} \]

Substitute back \(\theta = {\tan^{ - 1}}x\):

\[ u(x) = \frac{1}{2}{\tan^{ - 1}}x \]

Calculating the Derivatives with Respect to x

Now we need to find \(du/dx\) and \(dv/dx\).

For \(u(x) = \frac{1}{2}{\tan^{ - 1}}x\):

\[ \frac{{du}}{{dx}} = \frac{d}{{dx}}\left( {\frac{1}{2}{\tan^{ - 1}}x} \right) = \frac{1}{2} \cdot \frac{1}{{1 + {x^2}}} \]

For \(v(x) = {\tan^{ - 1}}x\):

\[ \frac{{dv}}{{dx}} = \frac{d}{{dx}}\left( {{\tan^{ - 1}}x} \right) = \frac{1}{{1 + {x^2}}} \]

Finding the Derivative of u with Respect to v

Finally, calculate \(\frac{{du}}{{dv}}\) using the formula \(\frac{{du/dx}}{{dv/dx}}\):

\[ \frac{{du}}{{dv}} = \frac{{\frac{1}{2} \cdot \frac{1}{{1 + {x^2}}}}}{{\frac{1}{{1 + {x^2}}}}} \]

Cancel out the common term \(\frac{1}{{1 + {x^2}}}\):

\[ \frac{{du}}{{dv}} = \frac{1}{2} \]

The derivative of \({\tan ^{ - 1}}\left( {\frac{{\sqrt {1{\rm{\;}} + {{\rm{x}}^2}} - {\rm{\;}}1}}{{\rm{x}}}} \right)\) with respect to \({\tan ^{ - 1}}{\rm{x}}\) is \(\frac{1}{2}\).

Revision Table: Derivatives of Inverse Trigonometric Functions

Additional Information: Differentiation using Substitution

Differentiation using substitution is a powerful technique, especially when dealing with complex functions involving roots, powers, or trigonometric expressions. The goal is to simplify the function into a form whose derivative is known or easier to calculate. In this problem, substituting \(x = \tan \theta\) inside the \({\tan ^{ - 1}}\) function simplified the complex expression significantly.

Steps for differentiation by substitution:

1. Identify a suitable substitution that simplifies the expression.
2. Rewrite the function in terms of the new variable.
3. Differentiate the simplified function with respect to the new variable (if needed, though here we differentiate with respect to x).
4. Substitute back the original variable.

In cases like the problem above, where we simplify \(u(x)\) first, the steps are slightly different:

1. Simplify the original function \(u(x)\) using a substitution (like \(x = \tan \theta\)) to get a simpler form in terms of x.
2. Once \(u(x)\) is simplified, differentiate \(u(x)\) with respect to x to find \(du/dx\).
3. Differentiate the second function \(v(x)\) with respect to x to find \(dv/dx\).
4. Calculate the desired derivative \(\frac{{du}}{{dv}} = \frac{{du/dx}}{{dv/dx}}\).

This method is particularly useful for simplifying inverse trigonometric functions containing algebraic expressions.