Question

$Differentiate space the space following space functions space by space substitutions space method space colon cos to the power of negative 1 end exponent open parentheses 2 straight x square root of 1 minus straight x squared end root close parentheses comma space minus fraction numerator 1 over denominator square root of 2 end fraction less than straight x less than fraction numerator 1 over denominator square root of 2 end fraction$

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Solution

Let space space straight y equals cos to the power of negative 1 end exponent open parentheses 2 straight x square root of 1 minus straight x squared end root close parentheses Put space space straight x equals sin space straight theta therefore space space space space straight y equals cos to the power of negative 1 end exponent open parentheses 2 sin space straight theta square root of 1 minus sin squared space straight theta end root close parentheses equals cos to the power of negative 1 end exponent open parentheses 2 sin space straight theta space cos space straight theta close parentheses space space space space space space space space space space equals cos to the power of negative 1 end exponent open parentheses sin space 2 straight theta close parentheses equals cos to the power of negative 1 end exponent open square brackets cos open parentheses straight pi over 2 minus 2 straight theta close parentheses close square brackets equals straight pi over 2 minus 2 straight theta therefore space space space space space straight y equals straight pi over 2 minus 2 space sin to the power of negative 1 end exponent straight x therefore space dy over dx equals 0 minus fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction equals negative fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction

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