Question 592647: (a) Find dy/dx for 
(b) Show that the point (x,y)= (3,1) lies on the curve defined by the equation in part (a), and find the slope of the tangent line at this point.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Take derivative of both sides with respect to x (note that since y is a function of x, you have to use the chain rule).
(4y^3)\frac{dy}{dx} + \frac{1}{2}(2x+3y)^{-\frac{1}{2}} (2 + 3 \frac{dy}{dx}) = 0)
Solve algebraically for dy/dx.
For part b, it is easy to show that (3,1) is on the curve because if you replace x=3, y=1 into the equation you should get a true statement. To find dy/dx at (3,1), replace x=3, y=1 into your expression for dy/dx (which you will have found in part a).
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