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Continuity And Differentiability

Question
CBSEENMA12035015
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FInd space dy over dx x equals straight a open parentheses cos space straight t plus log space tan straight t over 2 close parentheses comma space straight y equals straight a space sin space straight t.

Solution
space space space space space space space space space space space straight x equals straight a open parentheses cos space straight t plus tog space tan straight t over 2 close parentheses therefore space dx over dt equals straight a open square brackets negative sin space straight t plus fraction numerator 1 over denominator tan straight t over 2 end fraction. straight d over dt open parentheses tan straight t over 2 close parentheses close square brackets equals straight a open square brackets negative sin space straight t plus fraction numerator 1 over denominator tan straight t over 2 end fraction cross times sec squared straight t over 2 cross times 1 half close square brackets space space space space space space space space space space space space space equals straight a open square brackets negative sin space straight t plus fraction numerator cos begin display style straight t over 2 end style over denominator sin begin display style straight t over 2 end style end fraction cross times fraction numerator 1 over denominator cos squared begin display style straight t over 2 end style end fraction cross times 1 half close square brackets space space space space space space space space space space space space space equals straight a open square brackets negative sin space straight t plus fraction numerator 1 over denominator 2 space sin begin display style straight t over 2 end style cos begin display style straight t over 2 end style end fraction close square brackets equals straight a open square brackets negative sin space straight t plus fraction numerator 1 over denominator sin space straight t end fraction close square brackets equals straight a open square brackets fraction numerator negative sin squared space straight t plus 1 over denominator sin space straight t end fraction close square brackets equals fraction numerator straight a space cos squared straight t over denominator sin space straight t end fraction space space space space space space space space space space space straight y equals straight a space sin space straight t therefore space dy over dx equals straight a space cos space straight t 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 space cos space straight t cross times sin space straight t over denominator straight a space cos squared straight t end fraction equals fraction numerator sin space straight t over denominator cos space straight t end fraction equals tan space straight t

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