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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(x+Δx)=f(x)+f(x).Δxf(x)=logxf(x)=1xln10f(10.02))=f(10+0.02)=f(10)+f(10)×0.02=2.3026+110ln10×0.02=2.3026+21000ln10f(x)=cosxf(x)=sinxf(55)=f(54+1)=f(54)+f(54)×1=0.82930sin(54)=0.829301(0.82930)2

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