Question

Differentiate \(\tan^{-1}\left(\frac{\sqrt{1+x^2} - \sqrt{1-x^2}}{\sqrt{1+x^2} + \sqrt{1-x^2}}\right)\) with respect to \(\cos^{-1} x^2\).

Hint:

Choosing the right substitution (\(x^2 = \cos 2\theta\)) is the most important step for radical expressions involving \(1+x^2\) and \(1-x^2\).
Always simplify both functions in terms of the same parameter (\(\theta\)) to make the final division trivial.

Answer 1

Concept:
• Differentiation of one function \(u\) with respect to another \(v\) is given by \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\).
• Trigonometric substitutions help simplify complex inverse trig functions.

Step 1: Simplify the function \(u = \tan^{-1}\left(\dots\right)\)
Let \(x^2 = \cos 2\theta\). Then \(\theta = \frac{1}{2} \cos^{-1} x^2\).
Recall: \(1 + \cos 2\theta = 2\cos^2\theta\) and \(1 - \cos 2\theta = 2\sin^2\theta\).
So, \(\sqrt{1+x^2} = \sqrt{2}\cos\theta\) and \(\sqrt{1-x^2} = \sqrt{2}\sin\theta\).
Substituting into \(u\):
\[ u = \tan^{-1}\left(\frac{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}\right) = \tan^{-1}\left(\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta}\right) \] Divide numerator and denominator by \(\cos\theta\):
\[ u = \tan^{-1}\left(\frac{1 - \tan\theta}{1 + \tan\theta}\right) = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - \theta\right)\right] = \frac{\pi}{4} - \theta \]

Step 2: Simplify the function \(v = \cos^{-1} x^2\)
Since \(x^2 = \cos 2\theta\), we have:
\[ v = \cos^{-1}(\cos 2\theta) = 2\theta \]

Step 3: Calculate \(\frac{du}{dv}\)
We have \(u = \frac{\pi}{4} - \theta\) and \(v = 2\theta\).
Expressing \(\theta\) in terms of \(v\): \(\theta = \frac{v}{2}\).
Then, \(u = \frac{\pi}{4} - \frac{v}{2}\).
Differentiating \(u\) with respect to \(v\):
\[ \frac{du}{dv} = \frac{d}{dv}\left(\frac{\pi}{4} - \frac{v}{2}\right) = 0 - \frac{1}{2} = -\frac{1}{2} \]