question_answer

Using differentials find the approximate value of \[\sqrt{49.5}.\]

Answer:

Consider \[f\,(x)=\sqrt{x}\] \[\therefore \]      \[f'(x)=\frac{1}{2\sqrt{x}}\] Let \[x=49\]and \[\Delta \,x=0.5\] Now,     \[f\,(x+\Delta \,x)\approx f\,(x)+\Delta \,xf'\,(x)\] \[\Rightarrow \]   \[\sqrt{x+\Delta \,x}=\sqrt{x}+\frac{1}{2\sqrt{x}}x\,\Delta \,x\] \[\therefore \]\[\sqrt{49.5}\approx \sqrt{49}+\frac{0.5}{2\sqrt{49}}=7+\frac{0.5}{14}\approx 7+0.036\] \[\approx 7.036\]