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Continuity And Differentiability

Question
CBSEENMA12035162

Differentiate the following functions w.r.t.x: tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 minus cos space 2 straight x over denominator 1 plus cos space 2 straight x end fraction end root close parentheses

Solution
Let space straight y equals tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 1 minus cos space 2 straight x over denominator 1 plus cos space 2 straight x end fraction end root close parentheses equals tan to the power of negative 1 end exponent open parentheses square root of fraction numerator 2 sin squared straight x over denominator 2 cos squared straight x end fraction end root close parentheses equals tan to the power of negative 1 end exponent open parentheses fraction numerator sin space straight x over denominator cos space straight x 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 x right parenthesis
therefore space space space straight y equals straight x space rightwards double arrow space dy over dx equals 1

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