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Continuity And Differentiability

Question
CBSEENMA12035230
Wired Faculty App

Prove space that space dy over dx space is space independent space of space straight x comma space when straight y equals cot to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus sin space straight x end root plus square root of 1 minus sin space straight x end root over denominator square root of 1 plus sin space straight x end root minus square root of 1 minus sin space straight x end root end fraction close square brackets comma space open parentheses 0 less than straight x less than straight pi over 2 close parentheses.

Solution
space space space space straight y equals cot to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus sin space straight x end root plus square root of 1 minus sin space straight x end root over denominator square root of 1 plus sin space straight x end root minus square root of 1 minus sin space straight x end root end fraction close square brackets Now space square root of 1 plus sin space straight x end root equals square root of cos squared straight x over 2 plus sin squared straight x over 2 plus 2 sin straight x over 2 cos straight x over 2 end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals square root of open parentheses cos straight x over 2 plus sin straight x over 2 close parentheses squared end root equals cos straight x over 2 plus sin straight x over 2 square root of 1 minus sin space straight x end root equals square root of cos squared straight x over 2 plus sin squared straight x over 2 minus 2 sin straight x over 2 cos straight x over 2 end root equals square root of open parentheses cos straight x over 2 minus sin straight x over 2 close parentheses squared end root equals cos straight x over 2 minus sin straight x over 2 space space space space space therefore space space space space space space space space straight y equals cot to the power of negative 1 end exponent open parentheses fraction numerator open parentheses cos straight x over 2 plus sin straight x over 2 close parentheses plus open parentheses cos straight x over 2 minus sin straight x over 2 close parentheses over denominator open parentheses cos straight x over 2 plus sin straight x over 2 close parentheses minus open parentheses cos straight x over 2 minus sin straight x over 2 close parentheses end fraction close parentheses space space space space space space space space space space space space space space space space space space space equals cot to the power of negative 1 end exponent open parentheses fraction numerator 2 cos begin display style straight x over 2 end style over denominator 2 sin begin display style straight x over 2 end style end fraction close parentheses equals cot to the power of negative 1 end exponent open parentheses cot straight x over 2 close parentheses equals straight x over 2 space space space space space therefore space space dy over dx equals 1 half.

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