Question

Differentiate the following w.r.t.x: $tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 plus sin space straight x over denominator 1 minus sin space straight x end fraction end root close parentheses$

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Solution

Let space space space space space straight y equals tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 plus sin space straight x over denominator 1 minus sin space straight x end fraction end root close parentheses therefore space space space space space space straight y equals tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 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 over denominator 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 fraction end root close parentheses space space space space space space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator cos begin display style straight x over 2 end style plus sin begin display style straight x over 2 end style over denominator cos straight x over 2 minus sin straight x over 2 end fraction close parentheses equals tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator cos begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style end fraction plus fraction numerator sin begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style end fraction end style over denominator begin display style fraction numerator cos begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style end fraction minus fraction numerator sin begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style end fraction end style end fraction close square brackets space space space space space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 plus tan begin display style straight x over 2 end style over denominator 1 minus tan begin display style straight x over 2 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 4 plus straight x over 2 close parentheses close square brackets therefore space space space space space space space straight y equals straight pi over 4 plus straight x over 2 therefore space dy over dx equals 1 half

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Continuity And Differentiability

Differentiate the following w.r.t.x: $tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 plus sin space straight x over denominator 1 minus sin space straight x end fraction end root close parentheses$

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