y = cos^{-1} (frac{1-x^{2}}{1+x^{2}}), 0 < x < 1 में frac{d y}{d x} ज्ञात कीजिए।

Q: $y = \cos^{-1} (\frac{1-x^{2}}{1+x^{2}}), 0 < x < 1$ में $\frac{d y}{d x} $ ज्ञात कीजिए।

$\tan^{-1 }x = \theta$ अर्थात् $x = \tan \theta$ रखने पर, $\therefore y = \cos^{-1 } (\frac{1-x^{2}}{1+x^{2}}) = \cos^{-1} (\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}) (\because \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} = \cos 2\theta)$ $\Rightarrow y = \cos^{-1 }(\cos 2\theta) = 2\theta = 2 \tan^{-1} x$ $x$ के सापेक्ष अवकलन करने पर, $\frac{d y}{d x} = 2 \frac{d}{d x} (\tan^{-1} x) = \frac{2}{1+x^{2}} (\because \frac{d}{d x} \tan^{-1 }x = \frac{1}{1+x^{2}})$

प्रश्नों के उत्तर लिखिए। (प्रत्येक प्रश्न 3 अंक का हे)

Rajasthan - हिन्दी माध्यम

$y = \cos^{-1} (\frac{1-x^{2}}{1+x^{2}}), 0 < x < 1$ में $\frac{d y}{d x} $ ज्ञात कीजिए।

Exercise-5.3-11

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Solution

$\tan^{-1 }x = \theta$ अर्थात् $x = \tan \theta$ रखने पर,
$\therefore y = \cos^{-1 } (\frac{1-x^{2}}{1+x^{2}}) = \cos^{-1} (\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}) (\because \frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} = \cos 2\theta)$
$\Rightarrow y = \cos^{-1 }(\cos 2\theta) = 2\theta = 2 \tan^{-1} x$
$x$ के सापेक्ष अवकलन करने पर,
$\frac{d y}{d x} = 2 \frac{d}{d x} (\tan^{-1} x) = \frac{2}{1+x^{2}} (\because \frac{d}{d x} \tan^{-1 }x = \frac{1}{1+x^{2}})$

कक्षा 12 साइन्स
गणित
सांतत्य तथा अवकलनीयता
NCERT

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