Question

Using differentials, find the approximate value of the following up to 3 places of decimal.
$(401{)}^{\frac{1}{2}}$

Answer 1

$\left(401{)}^{\frac{1}{2}}Considerf\right(x)=\sqrt{x},\Rightarrow f^{\prime }(x)=\frac{1}{2\sqrt{x}}Letx=400and\Delta x=1Now,f(x+\Delta x)\simeq f(x)+\Delta x{f}^{\prime }(x)\Rightarrow \sqrt{x+\Delta x}\simeq \sqrt{x}+\frac{1}{2\sqrt{x}}\times \Delta x\Rightarrow \sqrt{400+1}\simeq \sqrt{400}+\frac{1}{2\sqrt{400}}=20+\frac{1}{2\times 20}=20+0.025=20.025\Rightarrow \sqrt{401}\simeq 20.025$