Using differentials, find the approximate value of 49.5

Question

Using differentials, find the approximate value of $\sqrt{49.5}$

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Answer

Let y = $\sqrt{x}\therefore\text{ y }+\Delta\text{y}=\sqrt{\text{x}+\Delta\text{x}}$
$ \Rightarrow\text{y}\frac{\text{dy}}{\text{dx}}.\Delta\text{x}\simeq\sqrt{\text{x}+\Delta\text{x}}$
$ \Rightarrow\sqrt{\text{x}}+\frac{\text{1}}{\text{2}\sqrt{x}}.\Delta\text{x}\simeq\sqrt{\text{x}+\Delta\text{x}}$
Putting x = 49 and Δx = 0.5 we get
$ \Rightarrow\sqrt{\text{49}}+\frac{\text{1}}{\text{2}\sqrt{49}}(0.5)\simeq\sqrt{\text{49.5}}$
$ \Rightarrow\sqrt{\text{49}}=7+\frac{\text{1}}{\text{28}}=\text{7.0357}$.

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