Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(0.0037{)}^{\frac{1}{2}}$

Answer 1

$\left(0.0037{)}^{\frac{1}{2}}Considerf\right(x)=\sqrt{x},\Rightarrow f^{\prime }(x)=\frac{1}{2\sqrt{x}}Letx=0.0036and\Delta x=0.0001Now,f(x+\Delta x)\simeq f(x)+\Delta x{f}^{\prime }(x)\Rightarrow \sqrt{0.0036+0.0001}=\sqrt{0.0036}+\frac{0.0001}{2\sqrt{0.0036}}=0.06+\frac{0.0001}{2\times 0.06}=0.06+0.0008=0.0608\Rightarrow \sqrt{0.0037}\simeq 0.0608$