(i) To find the gradient where the curve crosses the y-axis, substitute \(x = 0\) into the equation. The y-intercept is \(y = \frac{1 + 0^2}{1 + e^{0}} = \frac{1}{2}\).
Differentiate \(y = \frac{1 + x^2}{1 + e^{2x}}\) using the quotient rule:
\(\frac{d}{dx} \left( \frac{1 + x^2}{1 + e^{2x}} \right) = \frac{(1 + e^{2x}) \cdot 2x - (1 + x^2) \cdot 2e^{2x}}{(1 + e^{2x})^2}\).
Substitute \(x = 0\) into the derivative:
\(\frac{(1 + e^{0}) \cdot 0 - (1 + 0^2) \cdot 2e^{0}}{(1 + e^{0})^2} = \frac{-2}{4} = -\frac{1}{2}\).
(ii) For \(2x^3 + 5xy + y^3 = 8\), substitute \(x = 0\) to find \(y\):
\(2(0)^3 + 5(0)y + y^3 = 8 \Rightarrow y^3 = 8 \Rightarrow y = 2\).
Differentiate implicitly:
\(\frac{d}{dx}(2x^3) = 6x^2\), \(\frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx}\), \(\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\).
Combine to get:
\(6x^2 + 5y + 5x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0\).
Substitute \(x = 0\), \(y = 2\):
\(0 + 5(2) + 0 \cdot \frac{dy}{dx} + 3(2)^2 \frac{dy}{dx} = 0 \Rightarrow 10 + 12 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{5}{6}\).
