Let $y=\tan (5 x+2)$.
Find the differential $d y$ when $x=5$ and $d x=0.4$
Find the differential $d y$ when $x=5$ and $d x=0.8$

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Answer

$\boxed{dy = 4.0 + 4.0\tan^2(27)}$ when $x=5$ and $dx=0.8$.

Steps

Step 1 ：Given the function $y = \tan(5x + 2)$, we need to find the differential $dy$ when $x=5$ and $dx=0.4$, and when $x=5$ and $dx=0.8$.

Step 2 ：The differential of a function $y=f(x)$ is given by $dy=f'(x)dx$. So, we first need to find the derivative of the function $y=\tan (5x+2)$.

Step 3 ：The derivative of $y=\tan (5x+2)$ is $dy/dx = 5\sec^2(5x+2)$.

Step 4 ：Substitute $x=5$ into the derivative to get $dy/dx = 5\sec^2(27)$.

Step 5 ：Then, substitute $dx=0.4$ into the differential equation to get $dy = 5\sec^2(27) * 0.4$.

Step 6 ：Similarly, substitute $dx=0.8$ into the differential equation to get $dy = 5\sec^2(27) * 0.8$.

Step 7 ：Finally, we can simplify these expressions to get the final answers.

Step 8 ：$\boxed{dy = 2.0 + 2.0\tan^2(27)}$ when $x=5$ and $dx=0.4$.

Step 9 ：$\boxed{dy = 4.0 + 4.0\tan^2(27)}$ when $x=5$ and $dx=0.8$.