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Use differentials to find an approximate value for the given numbers $(1.97)^{6}$

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  • 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 $y=x^6$
Step 2:
let $x=2, \Delta x =-0.03$
Then $f(2)=2^6=64$
and $dy=6x^5dx$
$\quad= 6 \times 32 \times (-0.03)$
$\quad=-192 \times 0.03$
$\therefore f(1.97)=f(2)+\Delta y$
answered Aug 12, 2013 by meena.p

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