Question

Solve this:
Using the method of differential, find the approximate value of :
$\sqrt{37}$
(Ans: 6.083)

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^{1/2} \\ \mathrm{Let}\mathrm{x}=36\mathrm{and}\mathrm{x}+∆\mathrm{x}=37,\mathrm{then}∆\mathrm{x}=1 \\ \mathrm{For}\mathrm{x}=36,\mathrm{we}\mathrm{get} \\ \mathrm{y}={36}^{1/2}=6 \\ \mathrm{Let}\mathrm{dx}=∆\mathrm{x}=1 \\ \mathrm{Now},\mathrm{y}={\mathrm{x}}^{1/2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2{\mathrm{x}}^{1/2}} \\ \mathrm{Now},{\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{\mathrm{x}=36}=\frac{1}{2\times 6}=\frac{1}{12} \\ \mathrm{Now},\mathrm{dy}=\frac{\mathrm{dy}}{\mathrm{dx}}\times \mathrm{dx}=\frac{1}{12}\times 1=\frac{1}{12} \\ \mathrm{So},∆\mathrm{y}=\frac{1}{12} \\ \mathrm{Now},\sqrt{37}=\mathrm{y}+∆\mathrm{y}=6+\frac{1}{12}=\frac{73}{12}=6.083 \end{gathered}$$