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Continuity And Differentiability

Question
CBSEENMA12035259

Differentiate 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 space straight w. straight r. straight t. space sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses.

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

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