can i then cancel out the log(2x+1) from the top and one from the bottom leaving the denominator as log(2x+1) ?? thank u
so the ln(2x+1) dont cancel out ?
on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x
This seems a little bit misleading but maybe I am reading it poorly.
dy/dx(logx) is actually 1/x * 1/lnx, not 1/x * lnx which would put the lnx in the numerator instead of the denominator.