Differentiate the following functions with respect to `x` : `(e^x+sinx)/(1+logx)`

Explanation

Given,`(e^x+sinx)/(1+logx)`

Let, u=`(e^x+sinx)` and v=`(1+logx)`

`u'=(e^x+cosx)`

`v'=1/x`

now by formula

`(u/v)'`=`(u'v-v'u)/(v^2)`

=`((e^x+sinx)/(1+logx))'`=`[(e^x+cosx).(1+logx)-(e^x+sinx).(1/x)]/(1+logx)^2`

=`((e^x+sinx)/(1+logx))'`=`[x(e^x+cosx).(1+logx)-x(e^x+sinx).]/[x(1+logx)^2`]