**Q: **Find y'(3) using implicit differentiation of the equation:

2x^2 + 3x + xy = 3

and y(3) = -8

Answer:

First, take the derivative of the equation (this will involve a product rule on the “xy” term):

The implicit differentiation of **2x^2 + 3x + xy = 3** is:

4x + 3 + xy’ + y = 0

And we know that when x = 3, y = -8. So, plug that in and the solve for y’:

4(3) + 3 + 3y’ + (-8) = 0

12 + 3 + 3y’ – 8 = 0

7 + 3y’ = 0

3y’ = -7

y’ = -7/3

So, y'(3) = -7/3

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