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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. $(xii)\;(26.57)^{\Large\frac{1}{3}}$

This is twelveth part of multipart q1

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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 $y=x^{\Large\frac{1}{3}}$
$x=27$
$\Delta x=-0.43$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{3}}-(27)^{\Large\frac{1}{3}}$
$\quad=(26.57)^{\Large\frac{1}{3}}-(27)^{\Large\frac{1}{3}}$
$\quad=(26.57)^{\Large\frac{1}{3}}-3$
Step 2:
$(26.57)^{\Large\frac{1}{3}}=3+\Delta y$
$dy=\large\frac{dy}{dx}$$\Delta x=\large\frac{1}{3}x^{-\Large\frac{2}{3}}$$.\Delta x$
$\qquad\qquad\;\;=\large\frac{1}{3x^{\Large\frac{2}{3}}}$$\times \Delta x$
$\qquad\qquad\;\;=\large\frac{1}{3(27)^{\Large\frac{2}{3}}}$$\times (-0.43)$
$\qquad\qquad\;\;=\large\frac{1}{3(9)}$$\times (-0.43)$
$\qquad\qquad\;\;=-\large\frac{0.43}{27}$
$\qquad\qquad\;\;=-0.015926$
$\Delta y=-0.015926$
$\Rightarrow (26.57)^{\Large\frac{1}{3}}=3-0.015926$
$\qquad\;\;\;\qquad\;=2.984074$
$\qquad\;\;\;\qquad\;=2.984$(Approx)
answered Aug 5, 2013 by sreemathi.v
 

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