Given: \[ a(4 + x^2) = x + y - x^3 \] \[ x + y - x^3 = a^3 \frac{dy}{dx} \] Step 1: Implicitly differentiate the first equation with respect to \(x\): \[ \frac{d}{dx}[a(4 + x^2)] = \frac{d}{dx}[x + y - x^3] \Rightarrow a(2x) = 1 + \frac{dy}{dx} - 3x^2 \] Step 2: Rearrange to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = a(2x) - 1 + 3x^2 \quad \cdots (1) \] Step 3: Use the second original equation: \[ x + y - x^3 = a^3 \frac{dy}{dx} \quad \cdots (2) \] Substitute \( x = 1 \) into equation (1): \[ \frac{dy}{dx} = a(2 \cdot 1) - 1 + 3(1)^2 = 2a - 1 + 3 = 2a + 2 \quad \cdots (3) \] From the first original equation, at \( x = 1 \): \[ a(4 + 1^2) = 1 + y - 1^3 \Rightarrow a(5) = 1 + y - 1 \Rightarrow 5a = y \quad \cdots (4) \] Substitute into equation (2): \[ 1 + 5a - 1^3 = a^3 \cdot (2a + 2) \Rightarrow 5a = a^3(2a + 2) \] Assuming \( a eq 0 \), divide by \( a \): \[ 5 = a^2(2a + 2) \Rightarrow 5 = 2a^3 + 2a^2 \Rightarrow 2a^3 + 2a^2 - 5 = 0 \] Test the rational root \( a = 1 \) for the cubic equation: \[ 2(1)^3 + 2(1)^2 - 5 = 2 + 2 - 5 = -1 eq 0 \] For \( a = 1 \), \( \frac{dy}{dx} = 2a + 2 = 2(1) + 2 = 4 \)