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

To find the gradient of the curve, we need to differentiate the equation implicitly:

Starting with the equation:

$2x + 3y^2 + 3x^2y = 4x^2$

Differentiating both sides with respect to x:

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

This gives us:

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

Now substituting point P where $(x, y) = (-1, 1)$:

$2 + 6(1) \frac{dy}{dx} + 3(2(-1)(1) + (-1)^2 \frac{dy}{dx}) = 8(-1)$

Simplifying this leads to:

$2 + 6 \frac{dy}{dx} - 6 + 3 \frac{dy}{dx} = -8$

Combining like terms results in:

$9 \frac{dy}{dx} - 4 = -8$

Thus:

$9 \frac{dy}{dx} = -4 + 8 \Rightarrow 9 \frac{dy}{dx} = 4$

Finally, the gradient at P is:

$\frac{dy}{dx} = \frac{4}{9}$