a) Ta có:
\(\eqalign{
& y' = (cosx.{e^{2\tan x}})' = - \sin x{.e^{2\tan x}} \cr&\;\;\;\;\;\;\;+ \cos x.{2 \over {{{\cos }^2}x}}.{e^{2\tan x}} \cr
& = {e^{2\tan x}}({2 \over {\cos x}} - \sin x) \cr
& y' = ({\log _2}(\sin x))' = {{{\mathop{\rm cosx}\nolimits} } \over {\sin x}}.{1 \over {\ln 2}} = {{\cot x} \over {\ln 2}} \cr} \)
b) Ta có:
y’ = 4.e4x – 2e-x
y’’ = 16.e4x + 2e-x
y’’’ = 64.e4x – 2e-x
Suy ra: y’’’ – 13y’ – 12y = 64e4x – 2e-x – 13(4e4x - 2e-x ) – 12(e4x + 2e-x ) = 0
