Question

let ( y = 5sqrt{x} ).
find the change in ( y, delta y ) when ( x = 4 ) and ( delta x = 0.2 )
find the differential ( dy ) when ( x = 4 ) and ( dx = 0.2 )

Explanation:

Step1: Calculate \(\Delta y\)

The formula for \(\Delta y\) is \(y(x + \Delta x)-y(x)\).
Given \(y = 5\sqrt{x}\), when \(x = 4\) and \(\Delta x=0.2\), then \(x+\Delta x=4 + 0.2=4.2\).
\(y(4)=5\sqrt{4}=5\times2 = 10\), \(y(4.2)=5\sqrt{4.2}\approx5\times2.04939 = 10.24695\).
\(\Delta y=y(4.2)-y(4)=10.24695 - 10=0.24695\).

Step2: Calculate \(dy\)

First, find the derivative of \(y = 5\sqrt{x}=5x^{\frac{1}{2}}\).
Using the power rule \((x^n)^\prime=nx^{n - 1}\), \(y^\prime=\frac{5}{2}x^{-\frac{1}{2}}=\frac{5}{2\sqrt{x}}\).
The formula for \(dy=y^\prime dx\).
When \(x = 4\) and \(dx = 0.2\), \(y^\prime(4)=\frac{5}{2\sqrt{4}}=\frac{5}{4}\).
\(dy=\frac{5}{4}\times0.2=\frac{5\times0.2}{4}=\frac{1}{4}=0.25\).

Answer:

\(\Delta y = 0.24695\), \(dy=0.25\)