Question:
What is the derivative of \(\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) with respect to \(\tan^{-1} x\)?
\(\tan^{-1} x\) के सापेक्ष \(\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) का अवकलज क्या है?
Options:
- 0
- 1/2
- 1
- \(x\)
- 0
- 1/2
- 1
- \(x\)
Correct Answer: b)
Explanation:
Let \(u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) and \(v = \tan^{-1} x\).
Put \(x = \tan\theta\), so \(\theta = \tan^{-1} x = v\):
\[\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \frac{\sec\theta – 1}{\tan\theta} = \frac{1 – \cos\theta}{\sin\theta} = \tan\left(\frac{\theta}{2}\right)\]
Thus, \(u = \tan^{-1}\left(\tan\frac{\theta}{2}\right) = \frac{\theta}{2} = \frac{1}{2}v\).
Therefore, the derivative of \(u\) with respect to \(v\) is:
\[\frac{du}{dv} = \frac{1}{2}\]
Hence, the correct option is (b).
माना \(u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) तथा \(v = \tan^{-1} x\)।
\(x = \tan\theta\) रखने पर, \(\theta = v\):
\[\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta} = \frac{\sec\theta – 1}{\tan\theta} = \frac{1 – \cos\theta}{\sin\theta} = \tan\left(\frac{\theta}{2}\right)\]
अतः \(u = \frac{\theta}{2} = \frac{1}{2}v\)।
\(v\) के सापेक्ष अवकलन करने पर:
\[\frac{du}{dv} = \frac{1}{2}\]
अतः सही विकल्प (b) है।