Using implicit differentiation to find horizontal tangent?

Question

equation is x^3+y^3=3xy and I need to find the horizontal tangent line, thanks for any help

Answer 1

To solve by implicit differentiation, you need to solve for 'dy/dx'.

d[x^3]/dx + d[y^3]/dx = d[3xy]/dx. Taking the derivative, you get:

3x^2 + 3y^2(dy/dx) = 3y + 3x(dy/dx) [used the chain rule on the right side]. Isolating dy/dx, you get:

dy/dx = (3x^2 - 3y)/(3x - 3y^2). Factoring out a 3 in both the numerator and denominator, you get:

dy/dx = (x^2 - y)/(x - y^2)

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Now to solve for the horizontal tangent line, you need to determine when the slope (i.e. dy/dx) = 0. That is the definition of a horizontal line.

To solve dy/dx = 0, set the numerator = 0. Thus, you get:

x^2 = y

As you can see, there are an infinite number of ordered pairs (x,y) that satisfy this equation. Therefore, there are an infinite number of horizontal tangent lines that can be made. But make sure the denominator does not = 0, or the slope is undefined (vertical asymptote), or when x = y^2. An ordered pair that fits the above equation is (2,4). Therefore, y = 4 is a horizontal asymptote. y = 9 is a horizontal asymptote at (3,9).

The list goes on and on and on.....

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Hope this helps

Answer 2

Differentiate with respect to x:

3x^2 + 3y^2dy/dx = 3xdy/dx + 3y

Solve for dy/dx:

dy/dx = (3y - 3x^2)/(3y^2 - 3x) = 0 when y = x^2.

Now substitute y = x^2 into the original equation and solve for x:

x^3 + x^6 = 3x^3

x^6 - 2x^2 = x^3(x^3-2) = 0. The only solution of this equation that makes sense is x = 2^1/3, since dy/dx is undefined at x = 0, y = 0.

Therefore a horizontal tangent occurs at x = 2^(1/3), y = 2^(2/3), and the equation of the horizontal tangent is y = 2^(2/3)

Answer 3

Find dy/dx (or y') by the implicit differentiation method, then substitute dy/dx=0 since horizontal tangent lines have a slope of zero. Look at the set of points that satisfy the equation. You may want to solve for y in terms of x here, so that you can be descriptive about all of the points that satisfy this (so you have the x value and the y value of the point depends on the x value). Have fun.


Okay, if you combine all 3 answers you should get the jist of the problem (they tell you how to find the derivative, use my answer to find the points).

Answer 4

You need to differentiate both sides with respect to x. Don't forget to use the chain rule for all functions, where y is a function of x, as well as the multiplication rule on the right side.

x^3 + y^3 = 3xy
3x^2 + 3y^2(dy/dx) = 3x(dy/dx) + 3y
(dy/dx)(3y^2 - 3y) = -3x^2 + 3y
dy/dx = (-3x^2 + 3y) / (3y^2 - 3y)
dy/dx = (y - x^2) / (y^2 - 3y)

dy/dx is the slope of the horizontal tangent line. You need to set the expression for dy/dx equal to zero and solve it as a system with the equation of the line. The horzontal tangent has a slope of zero, and a y value that will be calculated when you solve the system. There may be more than one horizontal tangent line.

Answer 5

x^3+y^3=3xy
d/dx (x^3) + d/dx (y^3) = d/dx (3xy)
d/dx (x^3) + (d/dy (y^3)) (dy/dx) = d/dx (3xy)
3x^2 + 3y^2 (dy/dx) = 3y

For tangent, dy/dx = 0
hence
3x^2 + 0 = 3y
3x^2 = 3y
y = x^2

Answer 6

3x^2+3y^2y'=3xy'+3y
3(x^2-y)=3y'(x-y^2)
y'=(x^2-y)/(x-y^2)
when tgt is horizontal y'=0