let $y = \\tan(4x + 4)$. find the differential $dy$ when $x = 1$ and $dx = 0.3$ find the differential $dy$…

Explanation:

Step1: Find the derivative of y

The derivative of $y = \tan(u)$ with respect to $u$ is $\sec^{2}(u)$, and if $u = 4x + 4$, then by the chain - rule $\frac{dy}{dx}=\frac{d}{dx}\tan(4x + 4)=\sec^{2}(4x + 4)\cdot\frac{d}{dx}(4x + 4)$. Since $\frac{d}{dx}(4x + 4)=4$, we have $\frac{dy}{dx}=4\sec^{2}(4x + 4)$.

Step2: Recall the formula for the differential

The formula for the differential is $dy=\frac{dy}{dx}dx$.

Step3: Calculate dy when $x = 1$ and $dx = 0.3$

First, find $\sec^{2}(4x + 4)$ when $x = 1$. $4x+4=4\times1 + 4=8$, so $\sec^{2}(8)$. Then $dy=4\sec^{2}(8)\times0.3$. Since $\sec^{2}(8)=\frac{1}{\cos^{2}(8)}$, $dy = 1.2\sec^{2}(8)\approx1.2\times1.0142=1.21704$.

Step4: Calculate dy when $x = 1$ and $dx = 0.6$

Using $dy=\frac{dy}{dx}dx$ with $\frac{dy}{dx}=4\sec^{2}(4x + 4)$ and $x = 1$, $4x + 4=8$. Then $dy=4\sec^{2}(8)\times0.6=2.4\sec^{2}(8)\approx2.4\times1.0142 = 2.43408$.

Answer:

When $dx = 0.3$, $dy\approx1.21704$ When $dx = 0.6$, $dy\approx2.43408$