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Find the differential of the functions. $y$=$\sqrt{x^{4}+x^{2}+1}$

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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 $y=\sqrt {x^4+x^2+1}$
$dy=\large\frac{(4x^3+2x)dx}{2 {\sqrt {x^6+x^2+1}}}$
$\quad=\large\frac{x(2x^2+1)dx}{\sqrt {x^4+x^2+1}}$
answered Aug 12, 2013 by meena.p