Sure, I can guide you through the process step-by-step.
First, we start with the function [tex]\( y = \tan(4x + 4) \)[/tex].
To find the differential [tex]\( dy \)[/tex], we need to determine the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex], which is denoted as [tex]\( \frac{dy}{dx} \)[/tex]. This requires us to apply the chain rule.
Step-by-Step Solution:
1. Differentiate [tex]\( y = \tan(4x + 4) \)[/tex]:
Let [tex]\( u = 4x + 4 \)[/tex]. Then, [tex]\( y = \tan(u) \)[/tex].
The derivative of [tex]\( \tan(u) \)[/tex] with respect to [tex]\( u \)[/tex] is [tex]\( \sec^2(u) \)[/tex], and the derivative of [tex]\( u = 4x + 4 \)[/tex] with respect to [tex]\( x \)[/tex] is 4. Using the chain rule:
[tex]\[
\frac{dy}{dx} = \frac{d(\tan(u))}{du} \cdot \frac{du}{dx} = \sec^2(u) \cdot 4
\][/tex]
Substitute [tex]\( u = 4x + 4 \)[/tex]:
[tex]\[
\frac{dy}{dx} = 4 \sec^2(4x + 4)
\][/tex]
2. Evaluate [tex]\( \frac{dy}{dx} \)[/tex] at [tex]\( x = 4 \)[/tex]:
Substitute [tex]\( x = 4 \)[/tex] into [tex]\( u \)[/tex]:
[tex]\[
u = 4(4) + 4 = 16 + 4 = 20
\][/tex]
So,
[tex]\[
\frac{dy}{dx} \Big|_{x=4} = 4 \sec^2(20)
\][/tex]
3. Calculate [tex]\( dy \)[/tex] for [tex]\( dx = 0.4 \)[/tex] and [tex]\( dx = 0.8 \)[/tex]:
The differential [tex]\( dy \)[/tex] can be found using [tex]\( dy = \frac{dy}{dx} \cdot dx \)[/tex].
- For [tex]\( dx = 0.4 \)[/tex]:
[tex]\[
dy \Big|_{x=4, \, dx=0.4} = \left( 4 \sec^2(20) \right) \cdot 0.4 \approx 9.6078
\][/tex]
- For [tex]\( dx = 0.8 \)[/tex]:
[tex]\[
dy \Big|_{x=4, \, dx=0.8} = \left( 4 \sec^2(20) \right) \cdot 0.8 \approx 19.2156
\][/tex]
Summarizing the results:
- The differential [tex]\( dy \)[/tex] when [tex]\( x = 4 \)[/tex] and [tex]\( dx = 0.4 \)[/tex] is approximately [tex]\( 9.6078 \)[/tex].
- The differential [tex]\( dy \)[/tex] when [tex]\( x = 4 \)[/tex] and [tex]\( dx = 0.8 \)[/tex] is approximately [tex]\( 19.2156 \)[/tex].
Thus, these are the final differentials for the given values of [tex]\( dx \)[/tex].
