Question Number 65300 by rajesh4661kumar@gmail.com last updated on 28/Jul/19
Commented by kaivan.ahmadi last updated on 28/Jul/19
$${f}\left({x}+\Delta{x}\right)={f}\left({x}\right)+{f}'\left({x}\right).\Delta{x} \\ $$$${f}\left({x}\right)={logx}\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{{xln}\mathrm{10}} \\ $$$$\left.{f}\left(\mathrm{10}.\mathrm{02}\right)\right)={f}\left(\mathrm{10}+\mathrm{0}.\mathrm{02}\right)={f}\left(\mathrm{10}\right)+{f}'\left(\mathrm{10}\right)×\mathrm{0}.\mathrm{02}= \\ $$$$\mathrm{2}.\mathrm{3026}+\frac{\mathrm{1}}{\mathrm{10}{ln}\mathrm{10}}×\mathrm{0}.\mathrm{02}=\mathrm{2}.\mathrm{3026}+\frac{\mathrm{2}}{\mathrm{1000}{ln}\mathrm{10}} \\ $$$$ \\ $$$$ \\ $$$${f}\left({x}\right)={cosx}\Rightarrow{f}'\left({x}\right)=−{sinx} \\ $$$${f}\left(\mathrm{55}\right)={f}\left(\mathrm{54}+\mathrm{1}\right)={f}\left(\mathrm{54}\right)+{f}'\left(\mathrm{54}\right)×\mathrm{1}= \\ $$$$\mathrm{0}.\mathrm{82930}−{sin}\left(\mathrm{54}\right)=\mathrm{0}.\mathrm{82930}−\sqrt{\mathrm{1}−\left(\mathrm{0}.\mathrm{82930}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$