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# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(iii)\;\sqrt{0.6}$

This is third part of multipart q1

Toolbox:
• Let $y=f(x)$
• $\Delta x$ denote a small increment in $x$
• $\Delta y=f(x+\Delta x)-f(x)$
• $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
Let $y=\sqrt{0.6}$
$\Rightarrow y=\sqrt x$
$x=0.64$
$\Delta x=-0.04$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\quad\;\;=\sqrt{0.64+(-0.04)}-\sqrt{0.64}$
$\quad\;\;=\sqrt{0.6}-0.8$
$\sqrt{0.6}=\Delta y+0.8$
$\sqrt{0.6}=\Delta y+0.8$------(1)
Step 2:
$dy$ is the approximate value of $\Delta y$
$dy=\large\frac{dy}{dx}$$\times \Delta x \large\frac{dy}{dx}=\frac{1}{2\sqrt x}[Differentiating with respect to x] dy=\large\frac{1}{2\sqrt x}$$.(-0.04)$
$\quad=\large\frac{-0.04}{2\sqrt{0.64}}$
$\quad=\large\frac{-0.04}{2\times 0.8}$
$\quad=-0.025$
Step 3:
Substitute the value of $dy$ in equation (1)
$\sqrt{0.60}=0.8-0.025$
$\qquad\;\;=0.775$