Question

let ( y=\tan (2 x+7) ).
find the differential ( d y ) when ( x=2 ) and ( d x=0.2 )
find the differential ( d y ) when ( x=2 ) and ( d x=0.4 )

Explanation:

Step1: Find the derivative of \(y = \tan(2x + 7)\)

Using the chain rule, if \(y=\tan(u)\) and \(u = 2x+7\), then \(\frac{dy}{du}=\sec^{2}(u)\) and \(\frac{du}{dx}=2\). So \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2\sec^{2}(2x + 7)\)

Step2: Use the formula \(dy=\frac{dy}{dx}dx\)

When \(x = 2\), \(u=2x + 7=2\times2+7 = 11\). So \(\frac{dy}{dx}=2\sec^{2}(11)\)

Case 1: When \(dx = 0.2\)

\(dy=2\sec^{2}(11)\times0.2\)
Since \(\sec^{2}\theta=1+\tan^{2}\theta\), and \(\tan(11)\) is a value. But more simply, using the derivative formula directly \(dy = 2\sec^{2}(2x + 7)dx\). Substituting \(x = 2\) and \(dx=0.2\):
\(dy=2\sec^{2}(2\times2 + 7)\times0.2=2\sec^{2}(11)\times0.2\approx2\times(1+\tan^{2}(11))\times0.2\). Another way, since \(\frac{dy}{dx}=2\sec^{2}(2x + 7)\), when \(x = 2\), \(\frac{dy}{dx}\big|_{x = 2}=2\sec^{2}(11)\approx2\times(1 + (\tan(11))^{2})\approx2\times(1+(-0.99999)^{2})\approx2\). Then \(dy=\frac{dy}{dx}dx\approx2\times0.2 = 0.4\) (using the fact that \(\sec^{2}\theta=1+\tan^{2}\theta\) and \(\tan(11)\approx - 0.99999\), \(\sec^{2}(11)\approx1\))

Case 2: When \(dx = 0.4\)

Using \(dy=\frac{dy}{dx}dx\), with \(\frac{dy}{dx}\big|_{x = 2}\approx2\) (as above), then \(dy=2\times0.4=0.8\)

Answer:

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