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Continuity And Differentiability

Question
CBSEENMA12035231
Wired Faculty App

Differentiate colon space tan to the power of negative 1 end exponent open parentheses 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 parentheses space straight w. straight r. straight t. straight x.

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

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