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Use differentials to find an approximate value for the given numbers\[\]$y=3\sqrt{1.02}+4\sqrt{1.02}$

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
$3 \sqrt {1.02}+4 \sqrt {1.02}$
Step 1:
Let $y=x^{\large\frac{1}{3}}+x^{\large\frac{1}{4}}$
$dy=\bigg(\large\frac{1}{3}x^{\Large\frac{-2}{3}}+\large\frac{1}{4}x^{\Large\frac{-3}{4}}\bigg)$$dx$
$\quad=\bigg(\large\frac{1}{3x^{\Large\frac{-2}{3}}}+\large\frac{1}{4x^{\Large\frac{-3}{4}}}\bigg)$$dx$
Step 2:
Let $x=1$ and $dx \approx \Delta x=0.02$
Then $f(1)=1^{\large\frac{1}{3}}+1^{\large\frac{1}{4}}$$=2$
Also $dy=\bigg(\large\frac{1}{3}+\frac{1}{4}\bigg)$$(0.02)$
$\qquad=\large\frac{7}{12} $$(0.02)$
$\qquad=0.0117$
$\therefore f(1.02)=f(1)+dy$
$\qquad=2+0.0117=2.0117$
answered Aug 12, 2013 by meena.p
 

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