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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

1 Answer

  • 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 $y=\sqrt{0.6}$
$\Rightarrow y=\sqrt x$
$\Delta x=-0.04$
$\Delta y=\sqrt{x+\Delta x}-\sqrt x$
$\sqrt{0.6}=\Delta y+0.8$
$\sqrt{0.6}=\Delta y+0.8$------(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)$
$\quad=\large\frac{-0.04}{2\times 0.8}$
Step 3:
Substitute the value of $dy$ in equation (1)
answered Aug 5, 2013 by sreemathi.v

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