Question

let ( y=\tan (4 x + 7) ).
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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Explanation:

Step1: Find the derivative of \(y = \tan(4x + 7)\)

Using the chain rule, if \(y=\tan(u)\) and \(u = 4x+7\), then \(\frac{dy}{du}=\sec^{2}(u)\) and \(\frac{du}{dx}=4\). So \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=4\sec^{2}(4x + 7)\)

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

When \(x = 5\), \(u=4\times5 + 7=27\). So \(\frac{dy}{dx}\big|_{x = 5}=4\sec^{2}(27)\)

• For \(dx = 0.4\):

\(dy=4\sec^{2}(27)\times0.4\). Since \(\sec^{2}\theta=1+\tan^{2}\theta\), and \(\tan(27)\approx0.5095\), \(\sec^{2}(27)=1+(0.5095)^{2}\approx1 + 0.2596=1.2596\). Then \(dy=4\times1.2596\times0.4=2.01536\)

• For \(dx = 0.8\):

\(dy=4\sec^{2}(27)\times0.8\). Using \(\sec^{2}(27)\approx1.2596\), then \(dy=4\times1.2596\times0.8 = 4.03072\)

Answer:

When \(dx = 0.4\), \(dy\approx2.02\); when \(dx=0.8\), \(dy\approx4.03\)