Đáp án:

 `↓`

Giải thích các bước giải:

`a) y = (2^x +1)/(2^x -1)`

`y' = ((2^x +1)' .(2^x -1) -(2^x +1).(2^x -1)')/((2^x -1)^2)`

`y'= (2^x .ln2 . (2^x -1) -(2^x +1).2^x .ln2)/((2^x -1)^2)`

`y' = (2^x . ln2 .2^x -2^x .ln2 -2^x .ln2 .2^x - 2^x .ln2)/((2^x -1)^2)`

`y' = (-2.2^x .ln2)/((2^x -1)^2)`

`y' = (-2^x .ln4)/((2^x -1)^2)`

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`b) y = (3ln x +2)(2 log_3 x -5)`

`y' = (3ln x +2)' .(2 log_3 x -5)+(3ln x +2).(2 log_3 x -5)'`

`y'= 3/x .(2 log_3 x -5)+(3ln x +2). 2 . 1/(x ln3)`

`y' = (3.2 log_3 x)/x - (15)/x + (2.3ln x)/(x ln 3) + (2.2)/(x ln 3)`

`y' = (6 log_3 x)/x - (15)/x + (6lnx)/(x ln3) + 4/(x ln3)`