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Continuity And Differentiability

Question
CBSEENMA12035252
Wired Faculty App

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 2 straight x over denominator 1 minus straight x squared end fraction close parentheses

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

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