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Given xy + 3y + 2x^2= 7, find y' by implicit differentiation and by solving for y and show that the result is the same.    

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The expression xy + 3y + 2x^2= 7 has to be used to find the derivative y'

Using inplicit differentiation

y + x*y' + 3*y' + 4x = 0

=> y'(x + 3) = -y - 4x

=> y' = (-y - 4x)/(x + 3) ...(1)

Solving for y gives:

y(x + 3) = 7 - 2x^2

y = (7 - 2x^2)/(x + 3)

y' = (-4x*(x + 3) - (7 - 2x^2))/(x + 3)^2

=> (-4x^2 - 12x - 7 + 2x^2)/(x + 3)^2

=> (-4x^2 - 12x - y(x + 3))/(x + 3)^2

=> (-4x(x + 3) - y(x + 3))/(x + 3)^2

=> (-4x - y)/(x + 3) ...(2)

From (1) and (2) it can be seen that y' = (-4x - y)/(x + 3) irrespective of the method used.

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