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Consider the curve defined by 2y^3 +6x^2 y-12x^2+6y=1

a. Show that (dy/dx) = (4x-2xy)/((x^2)+(y^2)+1

b. Write an equation of each horizontal tangent to the curve.

c. The line throughthe orgin with slope -1 is tangent to the curve.at point P. Find the x- and y- coordinatrs of Point P.

Hi India,

This looks a lot like a homework question. I'll walk you thru part a) but please make an attempt at parts b)and c) in a follow-up question (I'll give a couple of hints).

a) To find dy/dx you need to perform so-called implicit differentiation. This basically says that if you want to take the derivative with respect to (wrt) x of an expression that contains a function of x, say y(x), which itself is contained in an algebraic expression, say f(y) ("function of a function", aka composite function), then

df/dx = df[y(x)] = (df/dy)(dy/dx).

Notice in the last term, the 'dy' components appear to cancel out, at least notationally. They don't really cancel out but the correspondance of the 1st and last term, where the 'dy' components seem to get hidden, is the reason behind calling it implicit differentiation.

As an example, d/dx [y^2] = 2y･dy/dy

Let me denote dy/dx = y', then

d/dx{ 2y^3 + 6x^2y - 12x^2 +6y = 1 ] = 6y^2y' + 12xy +6x^2y' - 24x + 6y' = 0.

Solving for y' gives

y' = 2x(2-y) / ( x^2 + y^2 + 1 )

as in your expression (with the missing parens on the right).

b) A horizontal tangent means a horizontal line passing thru the points where y' = 0, that is, where y(x) has slope = 0. Such a horizontal line would just be y = c = constant where y(c) has zero slope.

c) Need to find the point where y' = -1.

Good luck and let me know how you are doing.

Randy

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