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Continuity And Differentiability

Question
CBSEENMA12035097

Differentiate the following w.r.t.x:sin to the power of negative 1 end exponent left parenthesis straight x square root of straight x right parenthesis comma space 0 less or equal than straight x less or equal than 1

Solution
Let space space space space space straight y equals sin to the power of negative 1 end exponent left parenthesis straight x square root of straight x right parenthesis equals sin to the power of negative 1 end exponent open parentheses straight x to the power of 3 over 2 end exponent close parentheses
therefore space dy over dx equals fraction numerator 1 over denominator square root of 1 minus open parentheses straight x to the power of 3 over 2 end exponent close parentheses squared end root end fraction. straight d over dx open parentheses straight x to the power of 3 over 2 end exponent close parentheses equals fraction numerator 1 over denominator square root of 1 minus straight x cubed end root end fraction.3 over 2 straight x to the power of 1 half end exponent
therefore space dy over dx equals fraction numerator 3 square root of straight x over denominator 2 square root of 1 minus straight x cubed end root end fraction

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