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Continuity And Differentiability

Question

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$Find space dy over dx when space straight x equals straight a space sinθ comma space straight y equals straight a open parentheses cosθ plus log space tan straight theta over 2 close parentheses$

Solution

space space space space space space space space space space space straight x equals straight a space sinθ therefore space dx over dθ equals straight a space cosθ Also space space space space space space space space space space space straight y equals straight a open parentheses cosθ plus log space tan straight theta over 2 close parentheses therefore space dy over dθ equals straight a open square brackets negative sinθ plus fraction numerator 1 over denominator tan begin display style straight theta over 2 end style end fraction. sec squared straight theta over 2.1 half close square brackets space space space space space space space space space space space space space equals straight a open square brackets negative sinθ plus fraction numerator cos begin display style straight theta over 2 end style over denominator begin display style sin straight theta over 2 end style end fraction. fraction numerator 1 over denominator 2 cos squared begin display style straight theta over 2 end style end fraction close square brackets space space space space space space space space space space space space space equals straight a open square brackets negative sinθ plus fraction numerator 1 over denominator 2 sin begin display style straight theta over 2 end style cos begin display style straight theta over 2 end style end fraction close square brackets space space space space space space space space space space space space space equals straight a open square brackets negative sinθ plus 1 over sinθ close square brackets space space space space space space space space space space space space space equals straight a open parentheses fraction numerator negative sin squared straight theta plus 1 over denominator sinθ end fraction close parentheses equals fraction numerator straight a space cos squared straight theta over denominator sinθ end fraction Now space dy over dx equals fraction numerator begin display style dy over dθ end style over denominator begin display style dx over dθ end style end fraction equals fraction numerator straight a space cos squared straight theta over denominator sinθ end fraction cross times fraction numerator 1 over denominator straight a space cosθ end fraction equals cosθ over sinθ therefore space space space space space dy over dx equals cotθ

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