Giải thích các bước giải:
B1:
Ta có:
$\begin{array}{l}
y = \sqrt {x + \sqrt {1 + {x^2}} } \\
\Rightarrow y' = \dfrac{{1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}}}{{2\sqrt {x + \sqrt {1 + {x^2}} } }}\\
\Rightarrow 2\sqrt {1 + {x^2}} .y' = 2\sqrt {1 + {x^2}} .\dfrac{{1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}}}{{2\sqrt {x + \sqrt {1 + {x^2}} } }}\\
\Rightarrow 2\sqrt {1 + {x^2}} .y' = 2.\dfrac{{x + \sqrt {1 + {x^2}} }}{{2\sqrt {x + \sqrt {1 + {x^2}} } }}\\
\Rightarrow 2\sqrt {1 + {x^2}} .y' = \sqrt {x + \sqrt {1 + {x^2}} } \\
\Rightarrow 2\sqrt {1 + {x^2}} .y' = y
\end{array}$
B2:
Ta có:
$\begin{array}{l}
y = \cot 2x\\
\Rightarrow y' = 2.\dfrac{{ - 1}}{{{{\sin }^2}2x}} = \dfrac{{ - 2}}{{{{\sin }^2}2x}}
\end{array}$
Lại có:
$\begin{array}{l}
{y^2} + 1 = {\cot ^2}2x + 1 = \dfrac{1}{{{{\sin }^2}2x}}\\
\Rightarrow y' + 2{y^2} + 2 = \dfrac{{ - 2}}{{{{\sin }^2}2x}} + 2.\dfrac{1}{{{{\sin }^2}2x}} = 0\\
\Rightarrow y' + 2{y^2} + 2 = 0
\end{array}$
