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

Question

Using differentials, find the approximate values of the following:
$\sqrt{26}$

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Answer

Let $\text{x}=25,\\\text{x}+ \triangle\text{x}=26$$\triangle\text{x}=26- 25$
$=-1$
Let $\text{y}=\sqrt{\text {x}}$
$\frac{\text{dy}} {\text{dx}}=\frac{1}{2\sqrt{\text{x}}}$
$\Big(\frac{\text{dy}} {\text{dx}}\Big)_{\text{x}=25}=\frac{1} {2\sqrt{25}}$
$=\frac{1}{10}$
$=0.1$
$\triangle\text{y}= \Big(\frac{\text{dy}}{\text{dx}}\Big)_ {\text{x}=25}\times(\triangle\text{x})$
$=(0.01)(1)$
$=0.01$
$\sqrt{26}=\text{y}+ \triangle\text{y}$
$=\sqrt{\text {x}}+0.01$
$\sqrt{25}+0.1$
$\sqrt{26}=5.1$

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