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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. $(vii)\;(26)^{\Large\frac{1}{3}}$

This is seventh part of multipart q1

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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 $y=x^{\Large\frac{1}{3}}$
Also let $x=27$
$\Delta x=-1$
So that $x+\Delta x=26$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{3}}-x^{\Large\frac{1}{3}}$
$\quad=(26)^{\Large\frac{1}{3}}-(27)^{\Large\frac{1}{3}}$
$\quad=(26)^{\Large\frac{1}{3}}-3$
$(26)^{\Large\frac{1}{3}}=3+\Delta y$
Step 2:
$\large\frac{dy}{dx}=\frac{1}{3}$$x^{\Large\frac{-2}{3}}$
$\quad=\large\frac{1}{3x^{\Large\frac{2}{3}}}$
Now $\Delta y$ is approximately equal to $dy$
$dy=\big(\large\frac{dy}{dx}\big)$$\Delta x$
$\quad=\large\frac{1}{3x^{\Large\frac{2}{3}}}$$\Delta x$
$\quad=\large\frac{1}{3(27)^{\Large\frac{2}{3}}}$$\Delta x$
$\quad=\large\frac{1}{3(27)^{\Large\frac{2}{3}}}$$(-1)$
$\quad=\large\frac{-1}{3\times 9}$
$\quad=\large\frac{-1}{27}$
$\quad=-0.037037$
Step 3:
Approximate value of $\Delta y=dy=-0.037037$
Hence from (1) we get,
$(26)^{\Large\frac{1}{3}}=3-0.37037$
$\qquad\;\;=2.962963$
$\qquad\;\;=2.963$(approx)
answered Aug 5, 2013 by sreemathi.v
 

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