Toán Đạo hàm cấp 2013

C

conga222222

$\eqalign{
& {f^/}\left( x \right) = {1 \over x} \cr
& {f^{//}}\left( x \right) = - {1 \over {{x^2}}} \cr
& {f^{///}}\left( x \right) = - {2 \over {{x^3}}} \cr
& ... \cr
& {f^{\left( n \right)}}\left( x \right) = {{{{\left( { - 1} \right)}^{n + 1}}1*2*3*...*\left( {n - 1} \right)} \over {{x^n}}} = \cr} $
 
You must log in or register to reply here.
Top Bottom