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Continuity And Differentiability

Question
CBSEENMA12035254
Wired Faculty App

Differentiate space colon space cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus 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 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses

Solution
Let space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x squared over denominator 1 plus straight x squared end fraction close parentheses Put space space space space space straight x equals tan space straight theta therefore space space space space space space space straight y equals cos to the power of negative 1 end exponent open parentheses fraction numerator 1 minus tan squared space straight theta over denominator 1 plus tan squared space straight theta end fraction close parentheses space space space space space space space space space space space space space equals cos to the power of negative 1 end exponent left parenthesis cos space 2 straight theta right parenthesis equals 2 straight theta therefore space space space space space space space straight y equals 2 tan to the power of negative 1 end exponent space straight x therefore space space dy over dx equals fraction numerator 2 over denominator 1 plus straight x squared end fraction Also space space space space straight u equals 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 Put space space space space space space space straight x equals space tan space straight theta therefore space space space space space space space space space space space straight u equals space 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 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 equals 3 straight theta therefore space space space space space space space space space space space straight u equals 3 space tan to the power of negative 1 end exponent space straight x therefore space space space space space du over dx equals fraction numerator 3 over denominator 1 plus straight x squared 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 2 over denominator 1 plus straight x squared end fraction cross times fraction numerator 1 plus straight x squared over denominator 3 end fraction therefore space dy over dx equals 2 over 3

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