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Find the differential $dy$ and evaluate $dy$ for the given values of $x$ and $dx$\[\]$y$=$x^{4}-3x^{2}+x-1,x$$=2,dx$$=0.1$

This question has multiple parts. Therefore each part has been answered as a separate question on Clay6.com

1 Answer

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:
$y=x^4-3x^2+x-1$
$dy=(4x^3-9x^2+1)dx$
Step 2:
When $ x=2,dx=0.1$
$dy=[4(8)-9(4)+1]0.1=-0.3$
answered Aug 12, 2013 by meena.p
 

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