Using differentials, find the approximate values of the following… - Vidyadip

Question

Using differentials, find the approximate values of the following:$\frac{1}{\sqrt{25.1}}$

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Answer

Let $\text{x}=25,\text{x}+ \triangle\text{x}=251.$ $\triangle\text{x} =25.1-25$
$\triangle\text{x} =0.1$
Let $\text{y}=\frac{1} {\sqrt{\text{x}}}$
$\frac{\text{dy}} {\text{dx}}=\frac{2}{\text{x}^{\frac{3} {2}}}$
$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=25}=-\frac{1} {2(25)^{\frac{3}{2}}}$
$=-\frac{1}{250}$
$=-0.004$
Now,
$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=25}\times(\triangle\text{x})$
$=(-0.004)(0.1)$
$=-0.0004$
$\frac{1}{\sqrt{25.1}} =\text{y}+\triangle\text{y}$
$=\frac{1}{\sqrt{\text {x}}}+(-0.0004)$
$=\frac{1}{\sqrt{25}}- 0.0004$
$=\frac{1}{5}-0.0004$
$=0.2-0.0004$
$\frac{1}{\sqrt{25.1}} =0.1996$

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