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# Using differentials, find the approximate value of each of the following up to 3 places of decimal. $(ii)\;\sqrt{49.5}$

This is second 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 $\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\quad\;=\sqrt{49+.5}-\sqrt {49}$
$\quad\;=\sqrt{49.5}-\sqrt {49}$
$\quad\;=\sqrt{49.5}-7$
$\sqrt{49.5}=7+\Delta y$------(1)
Step 2:
$dy$ is the approximate value of $\Delta y$
$dy=\big(\large\frac{dy}{dx}\big)$$\Delta x We know that y=\sqrt x \large\frac{dy}{dx}=\frac{1}{2\sqrt x}[Differentiating with respect to x] dy=\large\frac{dy}{dx}$$.\Delta x$
$\quad\;=\large\frac{1}{2\sqrt x}$$.\Delta x \quad\;=\large\frac{1}{2\sqrt {49}}$$.0.5$
$\quad\;=\large\frac{1}{2\times 7}$$.0.5 \quad\;=\large\frac{1}{14}$$\times 0.5$
$\quad\;=0.036$
Step 3:
$\sqrt{49.5}=7+\Delta y$[From equation (1) we have]
Now substitute the value of $dy$ in (1)
$\sqrt{49.5}=7+0.036$
$\qquad\;\;=7.036$