find dy for the given values of x and δx. y = √(2x + 4); x = 2; δx = 0.2 dy = (round to the nearest…

Explanation:

Step1: Differentiate y with respect to x

First, rewrite $y = \sqrt{2x + 4}=(2x + 4)^{\frac{1}{2}}$. Using the chain - rule $\frac{dy}{dx}=\frac{1}{2}(2x + 4)^{-\frac{1}{2}}\times2=\frac{1}{\sqrt{2x + 4}}$.

Step2: Calculate dy

The differential $dy=\frac{dy}{dx}\Delta x$. Substitute $x = 2$ into $\frac{dy}{dx}$, we get $\frac{dy}{dx}\big|_{x = 2}=\frac{1}{\sqrt{2\times2+4}}=\frac{1}{\sqrt{8}}=\frac{1}{2\sqrt{2}}$. Then, since $\Delta x=0.2$, $dy=\frac{1}{2\sqrt{2}}\times0.2$. $dy=\frac{0.2}{2\sqrt{2}}=\frac{0.1}{\sqrt{2}}\approx\frac{0.1}{1.414}\approx0.071$.

Answer:

$0.071$