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x=8 dx=0,02
df=f'(x)·dx
∛7,98=∛8-1/12·0,02=2-0,01/6=2-0,0017=1,9983
sprawdzam 1,9983³=7,98 OK
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f(x)=ln(x) f'(x)=1/x
x=1 dx=0,05
f'(1)=1
ln(1,05)=ln(1)+1·0,05=0+0,05=0,05
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pozdr
Hans
Nalezy uwazac czy f. jest rosnaca czy malejaca !!!