Question
Download Solution PDFtan-1x के संबंध में \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) का अवकलज है:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
tan-1x के संबंध में \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) का अवकलज है:
अवधारणा:
g(x) के संबंध में f(x) का अवकलज \(\rm \frac{f'(x)}{g'(x)}\) है।
गणना:
माना \(f(x)=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\) और g(x) = tan-1x
फिर f(x) में x = tanθ रखने पर
\(\rm f(x)=\tan^{-1}\left(\frac{sec\theta-1}{tan\theta}\right)\)
\(\rm f(x)=\tan^{-1}\left(\frac{1-cos\theta}{sin\theta}\right)\)
\(\rm f(x)=\tan^{-1}\left(\frac{sin^2\frac{\theta}{2}}{2sin\frac{\theta}{2}cos\frac{\theta}{2}}\right)\)
\(\rm f(x)=\tan^{-1}\left(\tan\frac{\theta}{2}\right)\)
\(\rm f(x)=\frac{\theta}{2}\)
\(\rm f(x)=\frac12tan^{-1}x\)
अवकलित करने पर
\(\rm f'(x)=\frac{1}{2}\cdot\frac{1}{1+x^2}\)
हमारे पास g(x) = tan-1x है
अवकलित करने पर
\(\rm g'(x)=\frac{1}{1+x^2}\)
अब g(x) के संबंध में f(x) का अवकलज निम्न है
\(\rm \frac{f'(x)}{g'(x)}=\frac{\frac{1}{2}\cdot\frac{1}{1+x^2}}{\frac{1}{1+x^2}}=\frac{1}{2}\)
अतः विकल्प (2) सही है।
Last updated on Jun 19, 2025
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