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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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FInd the approximate value of $(1.999)^5$.

$\begin{array}{1 1} 31 \\ 31.92 \\ 30 \\ 31.82 \end{array} $

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1 Answer

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Toolbox:
  • If $ y = f(x)$, then $ \large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $ x $.
  • $ \bigg(\large \frac{dy}{dx} \bigg)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1
$(1.999)^5$
Let $ y=f(r)=(1.999)^5$
Let $x=2\: and \: x-\Delta x=1.999$
$ \therefore \Delta x=1.999+0.001=2$
$ \therefore x=2\: and\: \Delta x=0.001$
For $x=2$, we have $ y=(2)^5$
$=32$
$ dx=\Delta x=-0.001$
Step 2
$y=(x)^5$
Differentiating on both sides w.r.t $x$,
$ \large\frac{dy}{dx}=5x^4$
$ \Rightarrow dy=5x^4.dx$
When $x=2\: and \: dx=-0.001$
$dy=5(2)^4.(-0.001)$
$ = -80 \times 0.001$
$ = -.08 \Rightarrow \Delta y=-0.08$
Hence $(1.999)^5=y-\Delta y$
=$32-0.08$
$ = 31.92$
answered Jul 31, 2013 by thanvigandhi_1
 

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