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Continuity And Differentiability

Question
CBSEENMA12034918

Differentiate the following w.r.t.x: straight e to the power of square root of 1 minus straight x squared end root end exponent

Solution
Let space space space space space straight y equals straight e to the power of square root of 1 minus straight x squared end root end exponent. tan space straight x
therefore space dy over dx equals straight e to the power of square root of 1 minus straight x squared end root end exponent. straight d over dx left parenthesis tan space straight x right parenthesis plus tan space straight x. straight d over dx straight e to the power of square root of 1 minus straight x squared end root end exponent
space space space space space space space space space space space space space equals straight e to the power of square root of 1 minus straight x squared end root end exponent. sec squared straight x plus tan space straight x. straight e to the power of square root of 1 minus straight x squared end root end exponent. straight d over dx left parenthesis square root of 1 minus straight x squared end root right parenthesis
space space space space space space space space space space space space space equals straight e to the power of square root of 1 minus straight x squared end root end exponent sec squared straight x plus straight e to the power of square root of 1 minus straight x squared end root end exponent tan space straight x. open parentheses fraction numerator negative 2 straight x over denominator 2 square root of 1 minus straight x squared end root end fraction close parentheses equals straight e to the power of square root of 1 minus straight x squared end root end exponent open square brackets sec squared straight x minus fraction numerator straight x space tan space straight x over denominator square root of 1 minus straight x squared end root end fraction close square brackets

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