Help me ạ ????. Tại mình hết điểm r ạ
1. Cho hs y= x^3 – sin2x. Cm (y’-3x^2)^2 + (y”-6x / 2)^2 = 4
2. Cho hs y= √2x-x^2, với 0
Help me ạ ????. Tại mình hết điểm r ạ
1. Cho hs y= x^3 – sin2x. Cm (y’-3x^2)^2 + (y”-6x / 2)^2 = 4
2. Cho hs y= √2x-x^2, với 0
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
y = {x^3} – \sin 2x\\
\Rightarrow y’ = 3{x^2} – \left( {2x} \right)’.\cos 2x = 3{x^2} – 2\cos 2x\\
\Rightarrow y” = \left( {3{x^2} – 2\cos 2x} \right)’ = 3.2x – 2.\left( {2x} \right)’.\left( { – \sin 2x} \right) = 6x + 4\sin 2x\\
{\left( {y’ – 3{x^2}} \right)^2} + {\left( {\dfrac{{y” – 6x}}{2}} \right)^2}\\
= {\left( {3{x^2} – 2\cos 2x – 3{x^2}} \right)^2} + {\left( {\dfrac{{6x + 4\sin 2x – 6x}}{2}} \right)^2}\\
= {\left( { – 2\cos 2x} \right)^2} + {\left( {2\sin 2x} \right)^2}\\
= 4\left( {{{\cos }^2}2x + {{\sin }^2}2x} \right)\\
= 4.1 = 4\\
2.\\
y = \sqrt {2x – {x^2}} \\
\Rightarrow y’ = \dfrac{{\left( {2x – {x^2}} \right)’}}{{2\sqrt {2x – {x^2}} }} = \dfrac{{2 – 2x}}{{2\sqrt {2x – {x^2}} }} = \dfrac{{1 – x}}{{\sqrt {2x – {x^2}} }}\\
\Rightarrow y” = \dfrac{{\left( {1 – x} \right)’.\sqrt {2x – {x^2}} – \sqrt {2x – {x^2}} ‘.\left( {1 – x} \right)}}{{2x – {x^2}}}\\
= \dfrac{{ – 1.\sqrt {2x – {x^2}} – \dfrac{{1 – x}}{{\sqrt {2x – {x^2}} }}.\left( {1 – x} \right)}}{{2x – {x^2}}}\\
= \dfrac{{ – {{\sqrt {2x – {x^2}} }^2} – {{\left( {1 – x} \right)}^2}}}{{\left( {2x – {x^2}} \right)\sqrt {2x – {x^2}} }}\\
= \dfrac{{\left( {{x^2} – 2x} \right) – \left( {{x^2} – 2x + 1} \right)}}{{\left( {2x – {x^2}} \right)\sqrt {2x – {x^2}} }}\\
= \dfrac{{ – 1}}{{\left( {2x – {x^2}} \right)\sqrt {2x – {x^2}} }}\\
\Rightarrow {y^3}.y” + 1 = {\sqrt {2x – {x^2}} ^3}.\dfrac{{ – 1}}{{\left( {2x – {x^2}} \right)\sqrt {2x – {x^2}} }} + 1 = – 1 + 1 = 0
\end{array}\)