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Question-65300




Question Number 65300 by rajesh4661kumar@gmail.com last updated on 28/Jul/19
Commented by kaivan.ahmadi last updated on 28/Jul/19
f(x+Δx)=f(x)+f′(x).Δx  f(x)=logx⇒f′(x)=(1/(xln10))  f(10.02))=f(10+0.02)=f(10)+f′(10)×0.02=  2.3026+(1/(10ln10))×0.02=2.3026+(2/(1000ln10))      f(x)=cosx⇒f′(x)=−sinx  f(55)=f(54+1)=f(54)+f′(54)×1=  0.82930−sin(54)=0.82930−(√(1−(0.82930)^2 ))
$${f}\left({x}+\Delta{x}\right)={f}\left({x}\right)+{f}'\left({x}\right).\Delta{x} \\ $$$${f}\left({x}\right)={logx}\Rightarrow{f}'\left({x}\right)=\frac{\mathrm{1}}{{xln}\mathrm{10}} \\ $$$$\left.{f}\left(\mathrm{10}.\mathrm{02}\right)\right)={f}\left(\mathrm{10}+\mathrm{0}.\mathrm{02}\right)={f}\left(\mathrm{10}\right)+{f}'\left(\mathrm{10}\right)×\mathrm{0}.\mathrm{02}= \\ $$$$\mathrm{2}.\mathrm{3026}+\frac{\mathrm{1}}{\mathrm{10}{ln}\mathrm{10}}×\mathrm{0}.\mathrm{02}=\mathrm{2}.\mathrm{3026}+\frac{\mathrm{2}}{\mathrm{1000}{ln}\mathrm{10}} \\ $$$$ \\ $$$$ \\ $$$${f}\left({x}\right)={cosx}\Rightarrow{f}'\left({x}\right)=−{sinx} \\ $$$${f}\left(\mathrm{55}\right)={f}\left(\mathrm{54}+\mathrm{1}\right)={f}\left(\mathrm{54}\right)+{f}'\left(\mathrm{54}\right)×\mathrm{1}= \\ $$$$\mathrm{0}.\mathrm{82930}−{sin}\left(\mathrm{54}\right)=\mathrm{0}.\mathrm{82930}−\sqrt{\mathrm{1}−\left(\mathrm{0}.\mathrm{82930}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

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