Question

$If space straight x equals square root of straight a to the power of sin to the power of negative 1 end exponent straight t end exponent end root comma space straight y equals square root of straight a to the power of cos to the power of negative 1 end exponent straight t end exponent end root comma space show space that dy over dx equals negative straight y over straight x.$

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Solution

straight x equals open parentheses straight a to the power of sin to the power of negative 1 end exponent straight t end exponent close parentheses to the power of 1 half end exponent comma space straight y equals open parentheses straight a to the power of cos to the power of negative 1 end exponent straight t end exponent close parentheses to the power of 1 half end exponent therefore space log space straight x equals 1 half log space straight a to the power of sin to the power of negative 1 end exponent straight t end exponent equals 1 half sin to the power of negative 1 end exponent straight t. log space straight a equals 1 half loga space. sin to the power of negative 1 end exponent straight t Differentiating space both space sides space straight w. straight r. straight t. straight x comma space we space get comma space space space space space space space space 1 over straight x dx over dt equals 1 half log space straight a. fraction numerator 1 over denominator square root of 1 minus straight t squared end root end fraction therefore space space space space space space space space space dx over dt equals straight x open parentheses fraction numerator log space straight a over denominator 2 square root of 1 minus straight t squared end root end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis Again space log space straight y equals 1 half log open parentheses straight a to the power of cos to the power of negative 1 end exponent straight t end exponent close parentheses equals 1 half cos to the power of negative 1 end exponent straight t. log space straight a therefore space space space space space space space space log space straight y equals 1 half log space straight a. cos to the power of negative 1 end exponent straight t space space space space space space space 1 over straight y dy over dt equals 1 half log space straight a fraction numerator negative 1 over denominator square root of 1 minus straight t squared end root end fraction therefore space space space space space space space space space dy over dt equals negative straight y open parentheses fraction numerator log space straight a over denominator 2 square root of 1 minus straight t squared end root end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Now space space space space space dy over dx equals fraction numerator begin display style dy over dt end style over denominator begin display style dx over dt end style end fraction equals fraction numerator negative straight y open parentheses begin display style fraction numerator log space straight a over denominator 2 square root of 1 minus straight t squared end root end fraction end style close parentheses over denominator straight x open parentheses begin display style fraction numerator log space straight a over denominator 2 square root of 1 minus straight t squared end root end fraction end style close parentheses end fraction space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis left square bracket because space of space left parenthesis 1 right parenthesis comma left parenthesis 2 right parenthesis right square bracket therefore space space space space space space space space space space dy over dx equals negative straight y over straight x

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