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find instantaneous rate of change for function f(x)= log x, when x = 2.
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the answer is not 1/2.
what's the answer given
is it log to the base 10 or e
log base 10
?
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then change base first , we know that \[\log_{a} x= \log_{e} x/\log_{e} a\] so here \[\log_{10} x=\log_{e} x/\log_{e} 10\] so \[f(x)=\log_{e} x/\log_{e} 10\] now find f'(x) which is \[f'(x)=1/(x*\log_{e} 10)\] x=2 so \[f'(2)=1/(2*\log_{e} 10)\] \[\log_{e} 10=2.302\] \[f'(2)=1/(2*2.302)\] \[f'(2)=0.2172\]
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