cho y= log x+1 x. tính y'(1) . A. 0 b. 1/ln2
Giải thích
Đáp án B
y=logx+1x=lnxln(x+1)y'=1xln(x+1)-lnx.1x+1[ln(x+1)]2=(x+1)ln(x+1)-xlnxx(x+1)[ln(x+1)]2⇒y'(1)=2ln22ln22=1ln2
Đáp án B
y=logx+1x=lnxln(x+1)y'=1xln(x+1)-lnx.1x+1[ln(x+1)]2=(x+1)ln(x+1)-xlnxx(x+1)[ln(x+1)]2⇒y'(1)=2ln22ln22=1ln2