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Continuity And Differentiability

Question
CBSEENMA12034765
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If space straight x equals straight a open parentheses fraction numerator 1 plus straight t squared over denominator 1 minus straight t squared end fraction close parentheses space and space straight y equals fraction numerator 2 straight t over denominator 1 minus straight t squared end fraction comma space find space dy over dx.

Solution
Here space space straight x equals straight a open parentheses fraction numerator 1 plus straight t squared over denominator 1 minus straight t squared end fraction close parentheses space therefore space dx over dt equals straight a open square brackets fraction numerator left parenthesis 1 minus straight t squared right parenthesis. begin display style straight d over dt end style left parenthesis 1 plus straight t squared right parenthesis minus left parenthesis 1 plus straight t squared right parenthesis. begin display style straight d over dt end style left parenthesis 1 minus straight t squared right parenthesis over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction close square brackets space space space space space space space space space space space space space equals straight a open square brackets fraction numerator left parenthesis 1 minus straight t squared right parenthesis. left parenthesis 2 straight t right parenthesis minus left parenthesis 1 plus straight t squared right parenthesis. left parenthesis negative 2 straight t right parenthesis over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction close square brackets equals straight a open square brackets fraction numerator 2 straight t minus 2 straight t cubed plus 2 straight t plus 2 straight t cubed over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction close square brackets equals fraction numerator 4 at over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction space space space space space space space space space space space straight y equals fraction numerator 2 straight t over denominator 1 minus straight t squared end fraction therefore space dy over dt equals fraction numerator left parenthesis 1 minus straight t squared right parenthesis. begin display style straight d over dt end style left parenthesis 2 straight t right parenthesis minus left parenthesis 2 straight t right parenthesis. begin display style fraction numerator straight d over denominator dt left parenthesis 1 minus straight t squared right parenthesis end fraction end style over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction space space space space space space space space space space space space space equals fraction numerator left parenthesis 1 minus straight t squared right parenthesis.2 minus left parenthesis 2 straight t right parenthesis. begin display style left parenthesis negative 2 straight t right parenthesis end style over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction equals fraction numerator 2 minus 2 straight t squared plus 4 straight t squared over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction equals fraction numerator 2 plus 2 straight t squared over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction equals fraction numerator 2 left parenthesis 1 minus straight t squared right parenthesis over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction 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 2 left parenthesis 1 minus straight t squared right parenthesis over denominator left parenthesis 1 minus straight t squared right parenthesis squared end fraction cross times fraction numerator left parenthesis 1 minus straight t squared right parenthesis squared over denominator 4 at end fraction therefore space space space space space dy over dt equals fraction numerator 1 plus straight t squared over denominator 2 at end fraction

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