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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. $(xv)\;(32.15)^{\Large\frac{1}{5}}$

This is fifteenth 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 $y=x^{\Large\frac{1}{5}}$
$x=32$
$\Delta x=0.15$
$\Delta y=(x+\Delta x)^{\Large\frac{1}{5}}-x^{\Large\frac{1}{5}}$
$\quad=(32+0.15)^{\Large\frac{1}{5}}-(32)^{\Large\frac{1}{5}}$
$\quad=(32.15)^{\Large\frac{1}{5}}-2$
$\Rightarrow (32.15)^{\Large\frac{1}{5}}=2+\Delta y$------(1)
Step 2:
$\Delta y$ is the approximate value of $\Delta y$
$dy=\large\frac{dy}{dx}$$.\Delta x$
$\quad=\large\frac{1}{5}$$.x^{\Large\frac{-4}{5}}$$.\Delta x$
$\quad=\large\frac{1}{5x^{\Large\frac{4}{5}}}$$.\Delta x$
Step 3:
Substitute the value of $x$ and $\Delta x$
$\Rightarrow \large\frac{1}{5(32)^{\Large\frac{4}{5}}}$$\times (0.15)$
$\Rightarrow \large\frac{1}{5(2^5)^{\Large\frac{4}{5}}}$$\times (0.15)$
$\Rightarrow \large\frac{1}{5(2)^4}$$\times (0.15)$
$\Rightarrow \large\frac{1}{5\times 16}$$\times (0.15)$
$\Rightarrow \large\frac{0.15}{80}$
$\Rightarrow 0.001875$
Step 4:
Substitute the value of $\Delta x$ in equation (1)
$(32.15)^{\Large\frac{1}{5}}=2+\Delta y$
$\qquad\;\;\;\;\;\;\;=2+0.001875$
$\qquad\;\;\;\;\;\;\;=2.001875$
$\qquad\;\;\;\;\;\;\;=2.002$(Approx)
answered Aug 6, 2013 by sreemathi.v
 

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