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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).Δxf(x)=logx⇒f′(x)=1xln10f(10.02))=f(10+0.02)=f(10)+f′(10)×0.02=2.3026+110ln10×0.02=2.3026+21000ln10f(x)=cosx⇒f′(x)=−sinxf(55)=f(54+1)=f(54)+f′(54)×1=0.82930−sin(54)=0.82930−1−(0.82930)2
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