Certainly! Let's solve the given problem step-by-step for both [tex]\(\Delta y\)[/tex] and [tex]\(dy\)[/tex].
### 1. Finding the change in [tex]\( y \)[/tex], [tex]\(\Delta y\)[/tex]:
Given the function:
[tex]\[ y = 4x^2 \][/tex]
First, compute the initial value of [tex]\( y \)[/tex] when [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 4 \cdot (4)^2 = 4 \cdot 16 = 64 \][/tex]
Next, compute the new value of [tex]\( x \)[/tex] after the change, [tex]\(\Delta x\)[/tex]:
[tex]\[ \text{new } x = 4 + 0.4 = 4.4 \][/tex]
Now, compute the new value of [tex]\( y \)[/tex] with the updated [tex]\( x \)[/tex]:
[tex]\[ y_{\text{new}} = 4 \cdot (4.4)^2 \][/tex]
[tex]\[ (4.4)^2 = 19.36 \][/tex]
[tex]\[ y_{\text{new}} = 4 \cdot 19.36 = 77.44 \][/tex]
The change in [tex]\( y \)[/tex], [tex]\(\Delta y\)[/tex], is:
[tex]\[ \Delta y = y_{\text{new}} - y = 77.44 - 64 = 13.44 \][/tex]
### 2. Finding the differential [tex]\( dy \)[/tex]:
Given the differential form:
[tex]\[ dy = \frac{dy}{dx} \cdot dx \][/tex]
First, we find the derivative of the function [tex]\( y = 4x^2 \)[/tex]:
[tex]\[ \frac{dy}{dx} = 8x \][/tex]
Now compute [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\( x = 4 \)[/tex]:
[tex]\[ \frac{dy}{dx}\Bigg|_{x=4} = 8 \cdot 4 = 32 \][/tex]
Given [tex]\( dx = 0.4 \)[/tex], the differential [tex]\( dy \)[/tex] is:
[tex]\[ dy = 32 \cdot 0.4 = 12.8 \][/tex]
### Summary:
1. The change in [tex]\( y \)[/tex], [tex]\(\Delta y\)[/tex], is:
[tex]\[ \Delta y = 13.44 \][/tex]
2. The differential [tex]\( dy \)[/tex] is:
[tex]\[ dy = 12.8 \][/tex]
