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Find the differential of the functions.$y$=$4\sqrt{x}$

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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=4 \sqrt x$
$y=x^{\large\frac{1}{4}}$
$\therefore dy=x^{\Large\frac{-3}{4}}dx$
$\qquad= \large\frac{dx}{x^{\Large\frac{3}{4}}}$
answered Aug 12, 2013 by meena.p
 

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