The curve C has the equation 2x + 3y^2 + 3x^2y = 4x^2 - Edexcel - A-Level Maths Pure - Question 3 - 2012 - Paper 8

To find the gradient of the curve at point P, we need to perform implicit differentiation on the equation of the curve:

1. Differentiate the equation:

   $\frac{d}{dx}(2x) + \frac{d}{dx}(3y^2) + \frac{d}{dx}(3x^2y) = \frac{d}{dx}(4x^2)$

2. This gives us:

   $2 + 6y \frac{dy}{dx} + 3(2xy + x^2 \frac{dy}{dx}) = 8$

3. Rearranging, we get:

   $2 + 6y \frac{dy}{dx} + 6xy + 3x^2 \frac{dy}{dx} = 8$

4. Combine like terms:

   $(6y + 3x^2) \frac{dy}{dx} = 8 - 2 - 6xy$

5. Therefore, the gradient function is:

   $\frac{dy}{dx} = \frac{8 - 2 - 6xy}{6y + 3x^2}$

6. Substituting the coordinates of point P (-1, 1):

   $\frac{dy}{dx} \bigg|_{(-1, 1)} = \frac{8 - 2 - 6(-1)(1)}{6(1) + 3(-1)^2} = \frac{8 - 2 + 6}{6 + 3} = \frac{12}{9} = \frac{4}{3}.$

Thus, the gradient at P is $\frac{4}{3}$.