Find any stationary points on the curve 3x^2 +x^2y +y= 2?

Question

show working please

Answer 1

3x² + x²y + y = 2.------------------>(1).

Apply derivative each side.

(d/dx)(3x² + x²y + y) = (d/dx)2.

(d/dx)(3x²) + (d/dx)(x²y) + (d/dx)(y) = (d/dx)2.

Power Rule of Derivative: (d/dx)(xⁿ)=nx^n-1 and Product Rule: (UV)'=U'V+UV'

3(2x) + 2xy + x²y' + y' = 0

6x + 2xy + y'(x² +1) = 0

The stationary point on the curve is y' = 0 but y not equal to 0.

So, 6x + 2xy = 0

Take out common term x.

x(6 + 2y) = 0

Here x = 0

Substitute x = 0 in the equation (1).

3(0)² + (0)²y + y = 2

So, y = 2

Therefore the stationary point is (x, y)=(0, 2).