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

Question

Using differentials, find the approximate values of the following:
$\sqrt{37}$

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Answer

Consider the function $\text{y}=\text{f} (\text{x})=\sqrt{\text{x}}$
Let:
$\text{x} =36$
$\text{x}+\triangle \text{x}=37$
Then,
$\triangle\text{x}=1$
For $\text{x}=36$
$\text{y}=\sqrt{36} =6$
Let:
$\text{dx}=\triangle \text{x}=1$
Now $\text{y}=(\text{x})^ {\frac{1}{2}}$
$\Rightarrow\frac {\text{dy}}{\text{dx}}=\frac{1}{2\sqrt {\text{x}}}$
$\Rightarrow\Big(\frac {\text{dy}}{\text{dx}}\Big)_{\text{x} =36}=\frac{1}{12}$
$\therefore\triangle \text{y}=\text{dy}=\frac{\text{dy}} {\text{dx}}\text{dx}=\frac{1} {12}\times1=0.0833$
$\Rightarrow\triangle \text{y} =0.0833$
$\therefore\sqrt{37}= \text{y}+\triangle\text{y} =6.0833$

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