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Differential Calculus Applications - II
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Use differentials to find an approximate value for the given numbers $(1.97)^{6}$
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
q3-4
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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
Given $ (1.97)^6$
Step 1:
Let $y=x^6$
$dy=6x^5dx$
Step 2:
let $x=2, \Delta x =-0.03$
Then $f(2)=2^6=64$
and $dy=6x^5dx$
$\quad= 6 \times 32 \times (-0.03)$
$\quad=-192 \times 0.03$
$\quad=-5.76$
$\therefore f(1.97)=f(2)+\Delta y$
$\quad=64-5.76$
$\quad=58.24$
answered
Aug 12, 2013
by
meena.p
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