a)
\(\eqalign{
& y' = {{ - {{(x + 1)}'}} \over {{{(x + 1)}^2}}} = {{ - 1} \over {{{(x + 1)}^2}}} \cr
& \Rightarrow y'' = {{\left[ {{{(x + 1)}^2}} \right]'} \over {{{(x + 1)}^4}}} = {{2(x + 1)(x + 1)'} \over {{{(x + 1)}^4}}} \cr&\;\;\;\;\;\;\;\;\;\,= {2 \over {{{(x + 1)}^3}}} \cr} \)
b) Ta có: \(y = {1 \over x} + {1 \over {1 - x}}\)
Do đó:
\(\eqalign{
& y' = - {1 \over {{x^2}}} - {{(1 - x)'} \over {{{(1 - x)}^2}}} = - {1 \over {{x^2}}} + {1 \over {{{(1 - x)}^2}}} \cr
& y'' = {{({x^2})'} \over {{x^4}}} - {{\left[ {{{(1 - x)}^2}} \right]'} \over {{{(1 - x)}^4}}} \cr
& = {{2x} \over {{x^4}}} + {{2(1 - x)} \over {{{(1 - x)}^4}}} \cr
& = {2 \over {{x^3}}} + {2 \over {{{(1 - x)}^3}}} \cr} \)
c) \(y’ = (ax)’cos ax = a. cos ax\)
\(⇒ y’’ = -a (ax)’sin ax = -a^2sinax\)
d) \(y’ = 2sinx.(sinx)’ = 2sinx.cosx = sin 2x\)
\(⇒ y’’ = (2x)’.cos 2x = 2.cos 2x\)
