To find the gradient of the curve, we need to differentiate the given equation implicitly with respect to \(x\).
The equation is \(3e^{2x}y + e^xy^3 = 14\).
Differentiate each term:
1. Differentiate \(3e^{2x}y\) using the product rule:
\(\frac{d}{dx}(3e^{2x}y) = 3e^{2x} \cdot \frac{dy}{dx} + 6e^{2x}y\).
2. Differentiate \(e^xy^3\) using the product rule:
\(\frac{d}{dx}(e^xy^3) = e^xy^3 \cdot \frac{dy}{dx} + 3e^xy^2\).
Combine these results to get the derivative of the left-hand side:
\(6e^{2x}y + 3e^{2x} \frac{dy}{dx} + e^xy^3 \frac{dy}{dx} + 3e^xy^2 = 0\).
Substitute \(x = 0\) and \(y = 2\) into the equation:
\(6e^{0} \cdot 2 + 3e^{0} \cdot \frac{dy}{dx} + e^{0} \cdot 2^3 \cdot \frac{dy}{dx} + 3e^{0} \cdot 2^2 = 0\).
Simplify:
\(12 + 3 \frac{dy}{dx} + 8 \frac{dy}{dx} + 12 = 0\).
Combine like terms:
\(24 + 11 \frac{dy}{dx} = 0\).
Solve for \(\frac{dy}{dx}\):
\(11 \frac{dy}{dx} = -24\).
\(\frac{dy}{dx} = -\frac{24}{11}\).
Thus, the gradient of the curve at the point \((0, 2)\) is \(-\frac{4}{3}\).
