Let \( y=3 x^{2} \). Find the change in \( y, \Delta y \) when \( x=4 \) and \( \Delta x=0.1 \) Find the differential \( d y \) when \( x=4 \) and \( d x=0.1 \)

**Step 1. Compute \( y \) at \( x = 4 \) and \( x = 4.1 \)** Given \( y = 3x^2 \): - When \( x = 4 \): \[ y(4) = 3(4)^2 = 3 \times 16 = 48 \] - When \( x = 4.1 \): \[ y(4.1) = 3(4.1)^2 = 3 \times (16.81) = 50.43 \] **Step 2. Find the change in \( y \), denoted by \( \Delta y \)** \[ \Delta y = y(4.1) - y(4) = 50.43 - 48 = 2.43 \] **Step 3. Find the differential \( dy \)** First, find the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{d}{dx}(3x^2) = 6x \] At \( x = 4 \): \[ \frac{dy}{dx}\bigg|_{x=4} = 6 \times 4 = 24 \] Then, using \( dx = 0.1 \): \[ dy = \frac{dy}{dx}\bigg|_{x=4} \cdot dx = 24 \times 0.1 = 2.4 \] **Final Answers:** - The change in \( y \) is \( \Delta y = 2.43 \). - The differential is \( dy = 2.4 \).