Share
Giải giúp e với ạ e cần làm bài gấp ạ
Question
Leave an answer
Thu Thủy
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Answers ( )
$a)\quad y = (x+1)e^x$
$\to y’ = (x+1)’.e^x + (x+1).(e^x)’$
$\to y’ = e^x + (x+1)e^x$
$\to y’ = (x+2)e^x$
$b)\quad y = \dfrac{x^2 – x +2}{x+1}$
$\to y’=\dfrac{(x^2 – x +2)'(x+1) – (x^2 – x +2).(x+1)’}{(x+1)^2}$
$\to y’ =\dfrac{(2x-1)(x+1) – (x^2 – x +2)}{(x+1)^2}$
$\to y’ = \dfrac{x^2 + 2x -3}{(x+1)^2}$
$y=(x+1)e^x$
$y’=(x+1)’e^x+(x+1)(e^x)’$
$=e^x+(x+1)e^x$
$=e^x(x+2)$
$y=\dfrac{x^2-x+2}{x+1}$
$y’=\dfrac{(x^2-x+2)'(x+1)-(x^2-x+2)}{(x+1)^2}$
$=\dfrac{(2x-1)(x+1)-x^2+x-2}{(x+1)^2}$
$=\dfrac{2x^2+x-1-x^2+x-2}{(x+1)^2}$
$=\dfrac{x^2+2x-3}{(x+1)^2}$
$=\dfrac{(x-1)(x+3)}{(x+1)^2}$