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