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Differential Calculus Applications - II
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Use differentials to find an approximate value for the given numbers $\large\frac{1}{10.1}$
This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
tnstate
class12
bookproblem
ch6
sec-1
exercise6-1
p79
q3
q3-2
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asked
Apr 24, 2013
by
poojasapani_1
edited
Aug 12, 2013
by
meena.p
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1 Answer
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Toolbox:
Let $y=f(x)$ be a differentiable function then the quantities $dx$ and $dy$ are called differentials.The differential $dx$ is an independent variable.
The differential $dy$ is then defined by $dy=f'(x)dx(dx \approx \Delta x)$
Also $f(x+\Delta x)-f(x)=\Delta y =dy $ from which $f(x+\Delta x) $can be evaluated
$\large\frac{1}{10.1}=\frac{1}{10+0.1}$
Step 1:
Let $y=f(x)=\large\frac{1}{x}$
$dy=-\large\frac{1}{x^2}$$dx$
Step 2:
Let $x=10,dx \approx \Delta x=0.1$
Then $dy=-\large\frac{-1}{100}$$ \times 0.1=-0.001$
We also have $f(10)=\large\frac{1}{10}$$+dy$
Now $f(10.1)=f(10)+dy$
$\qquad\qquad=0.1-0.001$
$\qquad\qquad=0.099$
answered
Aug 12, 2013
by
meena.p
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