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Continuity And Differentiability

Question
CBSEENMA12034772
Wired Faculty App

Find space dy over dx when space straight x equals fraction numerator 3 at over denominator 1 plus straight t cubed end fraction comma space straight y equals fraction numerator 3 at squared over denominator 1 plus straight t cubed end fraction

Solution
space straight x equals fraction numerator 3 at over denominator 1 plus straight t cubed end fraction comma space straight y equals fraction numerator 3 at squared over denominator 1 plus straight t cubed end fraction therefore space dx over dt equals fraction numerator left parenthesis 1 plus straight t cubed right parenthesis begin display style straight d over dt end style left parenthesis 3 at right parenthesis minus 3 at begin display style straight d over dt end style left parenthesis 1 plus straight t cubed right parenthesis over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction space space space space space space space space space space space space space equals fraction numerator left parenthesis 1 plus straight t cubed right parenthesis.3 straight a minus 3 at. space 3 straight t squared over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction equals fraction numerator 3 straight a left parenthesis 1 minus 2 straight t cubed right parenthesis over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction and space dy over dt equals fraction numerator left parenthesis 1 plus straight t cubed right parenthesis begin display style straight d over dt end style left parenthesis 3 at squared right parenthesis minus 3 at squared begin display style straight d over dt end style left parenthesis 1 plus straight t cubed right parenthesis over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction space space space space space space space space space space space space space space space space equals fraction numerator left parenthesis 1 plus straight t cubed right parenthesis.6 at minus 3 at squared. space 3 straight t squared right parenthesis over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction equals fraction numerator 3 at left parenthesis 2 minus straight t cubed right parenthesis over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction therefore space dy over dt 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 3 at left parenthesis 2 minus straight t cubed right parenthesis over denominator left parenthesis 1 plus straight t cubed right parenthesis squared end fraction cross times fraction numerator left parenthesis 1 plus straight t cubed right parenthesis squared over denominator 3 straight a left parenthesis 1 minus 2 straight t cubed right parenthesis end fraction equals fraction numerator straight t left parenthesis 2 minus straight t cubed right parenthesis over denominator 1 minus 2 straight t cubed end fraction

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