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Find the differential $dy$ and evaluate $dy$ for the given values of $x$ and $dx$\[\]$y=\cos$$x,x=\large\frac{\pi}{6},$$dx=0.05$

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=\cos x $
$dy= -\sin x dx$
Step 2:
When $x=\large\frac{\pi}{6},$$ dx= 0.05$
$dy=-\sin \large\frac{\pi}{6} $$\times 0.05$
$\quad=\large\frac{-1}{2} $$\times 0.05$
$\quad=-0.025$
answered Aug 12, 2013 by meena.p
 

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