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Continuity And Differentiability

Question
CBSEENMA12034775
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For space straight a space positive space constant space straight a space find space dy over dx comma space where space straight y equals straight a to the power of straight t plus 1 over straight t end exponent comma space and space straight x equals open parentheses straight t plus 1 over straight t close parentheses to the power of straight a.

Solution
space space space space space space space space space space space space space space space straight x equals open parentheses straight t plus 1 over straight t close parentheses to the power of straight a comma space straight y equals straight a to the power of straight t plus 1 over straight t end exponent therefore space space space space space dx over dt equals straight a open parentheses straight t plus 1 over straight t close parentheses to the power of straight a minus 1 end exponent. straight d over dx open parentheses straight t plus 1 over straight t close parentheses therefore space space space space space dx over dt equals space straight a open parentheses straight t plus 1 over straight t close parentheses to the power of straight a minus 1 end exponent. open parentheses straight t minus 1 over straight t squared close parentheses equals straight a open parentheses straight t plus 1 over straight t close parentheses to the power of straight a minus 1 end exponent. open parentheses fraction numerator straight t squared minus 1 over denominator straight t squared end fraction close parentheses Also space dy over dt equals straight a to the power of straight t plus 1 over straight t end exponent log space straight a. straight d over dt open parentheses straight t plus 1 over straight t close parentheses equals straight a to the power of straight t plus 1 over straight t end exponent log space straight a. open parentheses 1 minus 1 over straight t squared close parentheses therefore space dy over dt equals straight a to the power of straight t plus 1 over straight t end exponent open parentheses fraction numerator straight t squared minus 1 over denominator straight t squared end fraction close parentheses log space straight a Now 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 straight a to the power of straight t plus 1 over straight t end exponent open parentheses fraction numerator straight t squared minus 1 over denominator straight t squared end fraction close parentheses log space straight a over denominator straight a open parentheses straight t plus 1 over straight t close parentheses to the power of straight a minus 1 end exponent open parentheses fraction numerator straight t squared minus 1 over denominator straight t squared end fraction close parentheses end fraction therefore space space space space space dy over dx equals fraction numerator straight a to the power of straight t plus 1 over straight t end exponent space log space straight a over denominator straight a open parentheses straight t plus 1 over straight t close parentheses to the power of straight a minus 1 end exponent end fraction.

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