Question

let ( y = \tan(2x + 7) ).
find the differential ( dy ) when ( x = 2 ) and ( dx = 0.2 )
find the differential ( dy ) when ( x = 2 ) and ( dx = 0.4 )
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Explanation:

Step1: Differentiate \(y = \tan(2x + 7)\)

Use the chain rule \((\tan u)^\prime=\sec^{2}u\cdot u^\prime\). Let \(u = 2x+7\), then \(u^\prime=2\). So \(y^\prime=\frac{dy}{dx}=2\sec^{2}(2x + 7)\)

Step2: Find \(dy\)

Since \(dy=y^\prime dx\), then \(dy = 2\sec^{2}(2x + 7)dx\)

Step3: Substitute \(x = 2\) and \(dx=0.2\)

First, find \(2x+7\) when \(x = 2\), \(2x+7=2\times2 + 7=11\). Then \(dy=2\sec^{2}(11)\times0.2\). Since \(\sec^{2}\theta=1+\tan^{2}\theta\), and \(\tan(11)\approx - 0.9999902\), \(\sec^{2}(11)=1+\tan^{2}(11)\approx1 + (-0.9999902)^{2}\approx2\). So \(dy=2\times2\times0.2 = 0.8\)

Step4: Substitute \(x = 2\) and \(dx = 0.4\)

Using \(dy=2\sec^{2}(2x + 7)dx\), with \(2x+7 = 11\) (when \(x = 2\)), and \(\sec^{2}(11)\approx2\). Then \(dy=2\times2\times0.4=1.6\)

Answer:

When \(dx = 0.2\), \(dy = 0.8\); when \(dx=0.4\), \(dy = 1.6\)