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Continuity And Differentiability

Question
CBSEENMA12035263
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Differentiate space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses space straight w. straight r. straight t. space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction close parentheses

Solution
Let space space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses Put space straight x equals tan space straight theta therefore space space space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator 3 tan space straight theta minus tan cubed space straight theta over denominator 1 minus 3 tan squared space straight theta end fraction close parentheses space space space space space space space space space equals tan to the power of negative 1 end exponent left parenthesis tan space 3 straight theta right parenthesis space space space space space space space space space equals 3 space straight theta therefore space space space straight y equals 3 space tan to the power of negative 1 end exponent straight x therefore space dy over dx equals fraction numerator 3 over denominator 1 plus straight x squared end fraction Also space straight u equals tan to the power of negative 1 end exponent space open parentheses fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction close parentheses Put space straight x equals sin space straight theta therefore space space space straight u equals tan to the power of negative 1 end exponent open parentheses fraction numerator sin space straight theta over denominator square root of 1 minus sin squared space straight theta end root end fraction close parentheses space space space space space space space space space equals tan to the power of negative 1 end exponent open parentheses fraction numerator sin space straight theta over denominator cos space straight theta end fraction close parentheses space space space space space space space space space equals tan to the power of negative 1 end exponent left parenthesis tan space straight theta right parenthesis space space space space space space space space space equals straight theta therefore space straight u equals sin to the power of negative 1 end exponent straight x therefore space du over dx equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction Now space dy over dx equals fraction numerator begin display style dy over dx end style over denominator begin display style du over dx end style end fraction equals fraction numerator 3 over denominator 1 plus straight x squared end fraction cross times fraction numerator square root of 1 plus straight x squared end root over denominator 1 end fraction equals fraction numerator 3 square root of 1 plus straight x squared end root over denominator 1 plus straight x squared end fraction

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