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Find the differential $dy$ and evaluate $dy$ for the given values of $x$ and $dx$\[\]$y=\sqrt{1-x},x=0,dx=0.02$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com
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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
Step 1:
$y=\sqrt {1-x}$
$dy=\large\frac{-1}{2 \sqrt {1-x}}$$dx$
Step 2:
When $x=0,dx=0.02$
$dy=\large\frac{-(0.02)}{2}$
$\qquad=-0.01$
answered Aug 12, 2013 by meena.p
 

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