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Continuity And Differentiability

Question
CBSEENMA12035267

Find space the space derivative space of space tan to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction space straight w. straight r. straight t. space tan to the power of negative 1 end exponent straight x

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

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