Đáp án:
\[P = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = x.\tan x\\
\Rightarrow y' = x'.\tan x + x.\left( {\tan x} \right)' = 1.\tan x + x.\dfrac{1}{{{{\cos }^2}x}}\\
= \tan x + x.\dfrac{1}{{{{\cos }^2}x}}\\
\Rightarrow y'' = \left( {\tan x} \right)' + x'.\dfrac{1}{{{{\cos }^2}x}} + x.\dfrac{{ - \left( {{{\cos }^2}x} \right)'}}{{{{\cos }^4}x}}\\
= \dfrac{1}{{{{\cos }^2}x}} + \dfrac{1}{{{{\cos }^2}x}} + x.\dfrac{{ - 2\left( {\cos x} \right)'.\cos x}}{{{{\cos }^4}x}}\\
= \dfrac{2}{{{{\cos }^2}x}} + x.\dfrac{{ - 2.\left( { - \sin x} \right).\cos x}}{{{{\cos }^4}x}}\\
= \dfrac{2}{{{{\cos }^2}x}} + \dfrac{{2x\sin x}}{{{{\cos }^3}x}}\\
P = {x^2}y'' - 2\left( {{x^2} + {y^2}} \right)\left( {1 + y} \right)\\
= {x^2}.\left( {\dfrac{2}{{{{\cos }^2}x}} + \dfrac{{2x\sin x}}{{{{\cos }^3}x}}} \right) - 2.\left( {{x^2} + {x^2}.{{\tan }^2}x} \right)\left( {1 + x\tan x} \right)\\
= \dfrac{{2{x^2}\cos x + 2{x^3}\sin x}}{{{{\cos }^3}x}} - 2.{x^2}\left( {1 + {{\tan }^2}x} \right)\left( {1 + x.\dfrac{{\sin x}}{{\cos x}}} \right)\\
= \dfrac{{2{x^2}\cos x + 2{x^3}\sin x}}{{{{\cos }^3}x}} - 2{x^2}.\dfrac{1}{{{{\cos }^2}x}}.\left( {1 + x.\dfrac{{\sin x}}{{\cos x}}} \right)\\
= \dfrac{{2{x^2}\cos x + 2{x^3}\sin x}}{{{{\cos }^3}x}} - \dfrac{{2{x^2}.\cos x + 2{x^3}\sin x}}{{{{\cos }^3}x}}\\
= 0
\end{array}\)
