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Continuity And Differentiability

Question
CBSEENMA12035205

Differentiate the following w.r.t.x: tan to the power of negative 1 end exponent left parenthesis square root of 1 plus straight x squared end root plus straight x right parenthesis

Solution
Let space straight y equals tan to the power of negative 1 end exponent left parenthesis square root of 1 plus straight x squared end root plus straight x right parenthesis semicolon space Put space straight x equals tan space straight theta
therefore space straight y equals tan to the power of negative 1 end exponent left parenthesis square root of 1 plus tan squared space straight theta end root plus tan space straight theta right parenthesis equals tan to the power of negative 1 end exponent left parenthesis sec space straight theta plus tan space straight theta right parenthesis
space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator cos space straight theta end fraction plus fraction numerator sin space straight theta over denominator cos space straight theta end fraction close parentheses space equals space tan to the power of negative 1 end exponent open parentheses fraction numerator 1 plus sin space straight theta over denominator cos space straight theta end fraction close parentheses
space space space space space space space space space equals space tan to the power of negative 1 end exponent space open square brackets fraction numerator cos squared begin display style straight theta over 2 end style plus sin space squared begin display style straight theta over 2 end style plus 2 sin begin display style straight theta over 2 end style cos begin display style straight theta over 2 end style over denominator cos squared begin display style straight theta over 2 end style minus sin squared begin display style straight theta over 2 end style end fraction close square brackets
space space space space space space space space space equals space tan to the power of negative 1 end exponent space open square brackets fraction numerator cos squared begin display style straight theta over 2 end style plus sin space squared begin display style straight theta over 2 end style plus 2 sin begin display style straight theta over 2 end style cos begin display style straight theta over 2 end style over denominator cos squared begin display style straight theta over 2 end style minus sin squared begin display style straight theta over 2 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator cos begin display style straight theta over 2 end style plus sin begin display style straight theta over 2 end style over denominator cos straight theta over 2 minus sin straight theta over 2 end fraction close square brackets
space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator cos begin display style straight theta over 2 end style plus sin begin display style straight theta over 2 end style over denominator cos straight theta over 2 minus sin straight theta over 2 end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 2 plus straight theta over 2 close parentheses close square brackets equals straight pi over 2 plus straight theta over 2 equals straight pi over 2 plus 1 half tan to the power of negative 1 end exponent straight x
therefore space dy over dx equals 0 plus 1 half. fraction numerator 1 over denominator 1 plus straight x squared end fraction space space rightwards double arrow space dy over dx equals fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction

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