Question

$If space space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x squared end root plus square root of 1 minus straight x squared end root over denominator square root of 1 plus straight x squared end root minus square root of 1 minus straight x squared end root end fraction close square brackets comma space prove space that space dy over dx equals negative fraction numerator straight x over denominator square root of 1 minus straight x to the power of 4 end root end fraction$

Solution

space space space space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus straight x squared end root plus square root of 1 minus straight x squared end root over denominator square root of 1 plus straight x squared end root minus square root of 1 minus straight x squared end root end fraction close square brackets Put space space straight x squared equals cos space straight theta therefore space space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus cos space straight theta end root plus square root of 1 minus cos space straight theta end root over denominator square root of 1 plus cos space straight theta end root minus square root of 1 minus cos space straight theta end root end fraction close square brackets equals tan space to the power of negative 1 end exponent open square brackets fraction numerator square root of 2 cos squared straight theta over 2 end root plus square root of 2 sin squared straight theta over 2 end root over denominator square root of 2 cos squared straight theta over 2 end root minus square root of 2 sin squared straight theta over 2 end root end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 2 space end root cos straight theta over 2 minus square root of 2 space end root sin space straight theta over 2 over denominator square root of 2 space end root cos straight theta over 2 plus square root of 2 space end root sin space straight theta over 2 end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator cos straight theta over 2 plus sin straight theta over 2 over denominator cos straight theta over 2 minus sin straight theta over 2 end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 plus tan space straight theta over 2 over denominator 1 minus tan space straight theta over 2 end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 4 plus straight theta over 2 close parentheses close square brackets equals straight pi over 4 plus straight theta over 2 therefore space space straight y equals straight pi over 4 plus 1 half cos to the power of negative 1 end exponent open parentheses straight x squared close parentheses therefore space dy over dx equals 0 plus 1 half open square brackets negative fraction numerator 1 over denominator square root of 1 minus straight x to the power of 4 end root end fraction close square brackets straight d over dx left parenthesis straight x squared right parenthesis equals 1 half cross times negative fraction numerator 1 over denominator square root of 1 minus straight x to the power of 4 end root end fraction cross times 2 straight x equals negative fraction numerator straight x over denominator square root of 1 minus straight x to the power of 4 end root end fraction.

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