Answer :
Let's work through this problem step-by-step:
1. Given Function:
[tex]\[ y = 2x^2 + 2x + 2 \][/tex]
2. Finding the Derivative:
To find the differential [tex]\( dy \)[/tex], we first need to determine the derivative of the given function [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]. The derivative [tex]\( \frac{dy}{dx} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 + 2x + 2) \][/tex]
Applying basic differentiation rules, we get:
[tex]\[ \frac{dy}{dx} = 4x + 2 \][/tex]
3. Evaluate the Derivative at [tex]\( x = 4 \)[/tex]:
Next, we substitute [tex]\( x = 4 \)[/tex] into the derivative to find the slope of the tangent line at that point:
[tex]\[ \frac{dy}{dx} \bigg|_{x=4} = 4(4) + 2 = 16 + 2 = 18 \][/tex]
4. Calculate [tex]\( dy \)[/tex] using the Differential Approximation [tex]\( dy \approx \frac{dy}{dx} \cdot dx \)[/tex]:
- For [tex]\( dx = 0.4 \)[/tex]:
[tex]\[ dy \approx \frac{dy}{dx} \cdot dx = 18 \cdot 0.4 = 7.2 \][/tex]
- For [tex]\( dx = 0.8 \)[/tex]:
[tex]\[ dy \approx \frac{dy}{dx} \cdot dx = 18 \cdot 0.8 = 14.4 \][/tex]
5. Summary of the Results:
- When [tex]\( x = 4 \)[/tex] and [tex]\( dx = 0.4 \)[/tex], the differential [tex]\( dy \)[/tex] is:
[tex]\[ dy = 7.2 \][/tex]
- When [tex]\( x = 4 \)[/tex] and [tex]\( dx = 0.8 \)[/tex], the differential [tex]\( dy \)[/tex] is:
[tex]\[ dy = 14.4 \][/tex]
Thus, we have found the required differentials for the given conditions.
1. Given Function:
[tex]\[ y = 2x^2 + 2x + 2 \][/tex]
2. Finding the Derivative:
To find the differential [tex]\( dy \)[/tex], we first need to determine the derivative of the given function [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]. The derivative [tex]\( \frac{dy}{dx} \)[/tex] is:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 + 2x + 2) \][/tex]
Applying basic differentiation rules, we get:
[tex]\[ \frac{dy}{dx} = 4x + 2 \][/tex]
3. Evaluate the Derivative at [tex]\( x = 4 \)[/tex]:
Next, we substitute [tex]\( x = 4 \)[/tex] into the derivative to find the slope of the tangent line at that point:
[tex]\[ \frac{dy}{dx} \bigg|_{x=4} = 4(4) + 2 = 16 + 2 = 18 \][/tex]
4. Calculate [tex]\( dy \)[/tex] using the Differential Approximation [tex]\( dy \approx \frac{dy}{dx} \cdot dx \)[/tex]:
- For [tex]\( dx = 0.4 \)[/tex]:
[tex]\[ dy \approx \frac{dy}{dx} \cdot dx = 18 \cdot 0.4 = 7.2 \][/tex]
- For [tex]\( dx = 0.8 \)[/tex]:
[tex]\[ dy \approx \frac{dy}{dx} \cdot dx = 18 \cdot 0.8 = 14.4 \][/tex]
5. Summary of the Results:
- When [tex]\( x = 4 \)[/tex] and [tex]\( dx = 0.4 \)[/tex], the differential [tex]\( dy \)[/tex] is:
[tex]\[ dy = 7.2 \][/tex]
- When [tex]\( x = 4 \)[/tex] and [tex]\( dx = 0.8 \)[/tex], the differential [tex]\( dy \)[/tex] is:
[tex]\[ dy = 14.4 \][/tex]
Thus, we have found the required differentials for the given conditions.
