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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: $ \left(\large\frac{17}{81}\right)^{\frac{1}{4}}$

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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)$$\Delta x$
Step 1:
$\big(\large\frac{17}{18}\big)^{\Large\frac{1}{4}}=\frac{(17)^{\Large\frac{1}{4}}}{(81)^{\Large\frac{1}{4}}}$
$\qquad\qquad=\large\frac{(17)^{\Large\frac{1}{4}}}{3}$
Let $f(x)=x^{\Large\frac{1}{4}}$
$\Rightarrow x=16$, $\Delta x=1$
$f(x+\Delta x)=(17)^{\Large\frac{1}{4}}$
$\Delta y=f(x+\Delta x)-f(x)$
$f(x+\Delta x)=f(x)+\Delta y$
Step 2:
$dy$ is approximately equal to $\Delta y$
$dy=\large\frac{dy}{dx}$$\times\Delta x$
$f(x)=x^{\Large\frac{1}{4}}$
Differentiating with respect to x we get
$f'(x)=\large\frac{1}{4}$$x^{\Large\frac{-3}{4}}$
$dy=\large\frac{1}{4}$$x^{\Large\frac{-3}{4}}$$\Delta x$
$\quad=\large\frac{1}{4}$$x^{\Large\frac{-3}{4}}$$.1$
$\Rightarrow \large\frac{1}{4(16)^{\Large\frac{3}{4}}}$$\times 1$
$\Rightarrow \large\frac{1}{4(8)}$$\times 1$
$\Rightarrow \large\frac{1}{32}$
Step 3:
$(17)^{\Large\frac{1}{4}}=2+\large\frac{1}{32}$$=2.03125$
$\big(\large\frac{17}{81}\big)^{\Large\frac{1}{4}}=\frac{(17)^{\Large\frac{1}{4}}}{3}$
$\qquad\;\;\;\;\;=\large\frac{2.03125}{3}$$=0.677083$
$\qquad\;\;\;\;\;=0.677$(Approx)
answered Aug 9, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

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