Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
y = x.\sin x\\
\Rightarrow y' = x'.\sin x + x.\left( {\sin x} \right)' = 1.\sin x + x.\cos x = \sin x + x.\cos x\\
\Rightarrow y'' = \left( {\sin x + x.\cos x} \right)' = \left( {\sin x} \right)' + x'.\cos x + x.\left( {\cos x} \right)'\\
= \cos x + 1.\cos x + x.\left( { - \sin x} \right) = 2\cos x - x.\sin x\\
\Rightarrow xy - 2\left( {y' - \sin x} \right) + x.y''\\
= x.x.\sin x - 2.\left( {\sin x + x.\cos x - \sin x} \right) + x.\left( {2\cos x - x.\sin x} \right)\\
= {x^2}\sin x - 2x.\cos x + 2x.\cos x - {x^2}\sin x\\
= 0\\
2,\\
y = \dfrac{{x - 3}}{{x + 4}}\\
\Rightarrow y' = \dfrac{{\left( {x - 3} \right)'.\left( {x + 4} \right) - \left( {x + 4} \right)'.\left( {x - 3} \right)}}{{{{\left( {x + 4} \right)}^2}}}\\
= \dfrac{{\left( {x + 4} \right) - \left( {x - 3} \right)}}{{{{\left( {x + 4} \right)}^2}}} = \dfrac{7}{{{{\left( {x + 4} \right)}^2}}} = 7.{\left( {x + 4} \right)^{ - 2}}\\
y'' = \left[ {7.{{\left( {x + 4} \right)}^{ - 2}}} \right]' = 7.\left( { - 2} \right).{\left( {x + 4} \right)^{ - 3}} = \dfrac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}}\\
2.{\left( {y'} \right)^2} - \left( {y - 1} \right)y''\\
= 2.{\left( {\dfrac{7}{{{{\left( {x + 4} \right)}^2}}}} \right)^2} - \left( {\dfrac{{x - 3}}{{x + 4}} - 1} \right).\dfrac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}}\\
= 2.\dfrac{{49}}{{{{\left( {x + 4} \right)}^4}}} - \left( {\dfrac{{ - 7}}{{x + 4}}} \right).\dfrac{{ - 14}}{{{{\left( {x + 4} \right)}^3}}}\\
= \dfrac{{98}}{{{{\left( {x + 4} \right)}^4}}} - \dfrac{{98}}{{{{\left( {x + 4} \right)}^4}}} = 0\\
\Rightarrow 2.{\left( {y'} \right)^2} = \left( {y - 1} \right)y''
\end{array}\)
