$\displaystyle y = \frac{e^xlogx}{log(2x+1)}$

$\displaystyle

y '= \frac{e^x(logx+\frac{1}{x})log(2x+1)-(\frac{2}{2x+1})(e^xlogx)}{(log(2x+1))^2}

$

can i thencancel outthe log(2x+1) from the top and one from the bottom leaving the denominator as log(2x+1) ?? thank u