Question

let ( y = 4x^{2} ).
find the change in ( y ), ( delta y ) when ( x = 1 ) and ( delta x = 0.1 )
find the differential ( dy ) when ( x = 1 ) and ( dx = 0.1 )
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Explanation:

Step1: Calculate \(\Delta y\)

The formula for \(\Delta y\) is \(y(x + \Delta x)-y(x)\).
Given \(y = 4x^{2}\), when \(x = 1\) and \(\Delta x=0.1\), \(x+\Delta x=1 + 0.1=1.1\).
\(y(1.1)=4\times(1.1)^{2}=4\times1.21 = 4.84\), \(y(1)=4\times1^{2}=4\).
\(\Delta y=y(1.1)-y(1)=4.84 - 4\).

Step2: Calculate \(dy\)

First, find the derivative of \(y = 4x^{2}\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(y^\prime=\frac{dy}{dx}=8x\).
The formula for \(dy\) is \(dy=y^\prime(x)dx\).
When \(x = 1\) and \(dx = 0.1\), \(y^\prime(1)=8\times1 = 8\).
\(dy=y^\prime(1)\times dx=8\times0.1\).

Answer:

\(\Delta y = 0.84\)
\(dy=0.8\)