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Continuity And Differentiability

Question
CBSEENMA12034754

Find space dy over dx space when space straight y equals straight v to the power of 4 over 4 space and space straight v equals 2 over 3 straight x cubed plus 7

Solution
Here space straight y equals straight v to the power of 4 over 4 comma straight v equals 2 over 3 straight x cubed plus 7 equals 1 third left parenthesis 2 straight x cubed plus 21 right parenthesis
therefore space space space space space space straight y equals 1 fourth open square brackets 1 third left parenthesis 2 straight x cubed plus 21 right parenthesis close square brackets to the power of 4 equals 1 over 324 left parenthesis 2 straight x cubed plus 21 right parenthesis to the power of 4
therefore space dy over dx equals 1 over 324.4 left parenthesis 2 straight x cubed plus 21 right parenthesis cubed. straight d over dx left parenthesis 2 straight x cubed plus 21 right parenthesis
space space space space space space space space space space space space space equals 1 over 81 left parenthesis 2 straight x cubed plus 21 right parenthesis. left parenthesis 6 straight x squared plus 0 right parenthesis
space space space space space space space space space space space space space equals fraction numerator 2 straight x squared left parenthesis 2 straight x cubed plus 21 right parenthesis cubed over denominator 27 end fraction

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