For Example, to Differentiate Y3, We Do It Like This

4.Consider the curve defined by 2y3+6x2y-12x2+6y = 1.
(a)Find dy/dx
(b)Write an equation of each horizontal tangent line to the curve.
(c)The line through the origin with slope –1 is tangent to the curve at point P. Find the x- and y- coordinates of point P.

(a)Find dy/dx

The eqn is: 2y3+6x2y-12x2+6y = 1

It would be impossible to get y in terms of x only from this eqn, so we differentiate the eqn implicitly.

For example, to differentiate y3, we do it like this,

And we use the product rule on terms like x2y.

When we carry out the implicit differentiation, this gives us,

2.3y2.y’ + 6(2xy + x2y’) – 12.2x + 6y’ = 0

6y2y’ + 12xy + 6x2y’ – 24x + 6y’ = 0

y’(6y2 + 6x2 + 6) + 12xy – 24x = 0

6y’(y2 + x2 + 1) = 12x(2 – y)

y’ = 2x(2 – y)/(y2 + x2 + 1)

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

When y’ = 0, then the numerator in the expression for y’ must be zero. Thus,

2x(2 – y) = 0

x = 0, y = 2

At x = 0, substitute for x = 0 into 2y3+6x2y-12x2+6y = 1, giving

2y3+0-0+6y = 1

y(y2 +3) = 0

y = 0, (only one real root, the others are imaginary)

The tangent line here is at (x,y) = (0, 0).

The eqn of this tangent line is therefore: y = 0. (the x-axis)

The other tangent line is given by: y = 2.

(c)The line through the origin with slope –1 is tangent to the curve at point P. Find the x- and y- coordinates of point P

When the slope = -1, we can put y’ = -1 in the eqn for the slope. So,

-1 = 2x(2 – y)/ (y2 + x2 + 1)

-(y2 + x2 + 1) = 2x(2 – y)

-y2 – x2 – 1 = 4x – 2xy

x2 + y2 + 4x – 2xy + 1 = 0

Since the coordinates of P lie on the line y = -x (the eqn with slope –1 passing thro’ the origin), then we can substitute for y = -x into the above quadratic eqn.

x2 + x2 + 4x + 2x2 + 1 = 0

4x2 + 4x + 1 = 0

(2x + 1)2 = 0

x = -½

y = -x = ½

The coordinates of P are: (- ½ , ½ ).

Fermat.