Question
Download Solution PDFयदि f(x) = \(\rm \sin^{-1}(1-x^2)\) है, तो f'(x) का मान ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
\(\rm d\over dx\) sin-1 x = \(\rm 1\over\sqrt{1 - x^2}\)
\(\rm d\over dx\) tan-1 x = \(\rm 1\over1 + x^2\)
श्रृंखला नियम (प्रतिस्थापन द्वारा अवकलन): यदि y, u का फलन है और u, x का फलन है।
\(\rm {dy\over dx} = {dy\over du}× {du\over dx}\)
गणना:
माना कि u = 1 - x2 है।
\(\rm du \over dx\) = - 2x
y = sin-1(1 - x2) = sin-1 u
\(\rm dy\over dx\) = \(\rm {dy\over du}×{du\over dx}\)
\(\rm dy\over dx\) = \(\rm d\over du\) sin-1 u × (-2x)
\(\rm dy\over dx\) = \(\rm 1\over\sqrt{1 - u^2}\) × (-2x)
\(\rm dy\over dx\) = \(\rm -2x\over\sqrt{1 - (1-x^2)^2}\)
\(\rm dy\over dx\) = \(\rm -2x\over\sqrt{2x^2-x^4}\)
\(\rm dy\over dx\) = \(\boldsymbol{\rm -2\over\sqrt{2-x^2}}\)
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