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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Using differentials, find the approximate value of each of the following up to $3$ places of decimal. $(3.968)^{\Large\frac{3}{2}}$

$\begin{array}{1 1} 2.904 \\ 7.904 \\ 4.904 \\ 5.904 \end{array} $

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1 Answer

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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 $y=x^{\Large\frac{3}{2}}$
$x=4$
$\Delta x=-0.032$
$\Delta y=(x+\Delta x)^{\Large\frac{3}{2}}-x^{\Large\frac{3}{2}}$
$\quad\;\;=(4-0.032)^{\Large\frac{3}{2}}-4^{\Large\frac{3}{2}}$
$\quad\;\;=(3.968)^{\Large\frac{3}{2}}-8$
$(3.968)^{\Large\frac{3}{2}}$$=8+\Delta y$------(1)
Step 2:
$dy$ is the approximate value of $\Delta y$
$dy=\large\frac{dy}{dx}$$\times \Delta x$
$\quad=\large\frac{3}{2}x^{\Large\frac{1}{2}}.\Delta x$
$\quad=\large\frac{3}{2}$$\sqrt x.\Delta x$
$\Rightarrow \large\frac{3}{2}$$\sqrt 4\times (-0.032)$
$\Rightarrow \large\frac{3}{2}$$2\times (-0.032)$
$\Rightarrow 3\times (-0.032)$
$\Rightarrow -0.096)$
Step 3:
From (1) we have
$(3.968)^{\Large\frac{3}{2}}=8+\Delta y$
$\qquad\;\;\;\;\;\;\;=8-0.096$
$\qquad\;\;\;\;\;\;\;=7.904$
answered Aug 5, 2013 by sreemathi.v
 

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