Các quy tắc tính đạo hàm

Bước 1:

$y' = \dfrac{{\left( {\sin 2x + 2} \right)'\left( {\cos 2x + 3} \right) - \left( {\cos 2x + 3} \right)'\left( {\sin 2x + 2} \right)}}{{{{\left( {\cos 2x + 3} \right)}^2}}}$

Bước 2:

$\begin{array}{l}y' = \dfrac{{2cox2x\left( {\cos 2x + 3} \right) + 2\sin 2x\left( {\sin 2x + 2} \right)}}{{{{\left( {\cos 2x + 3} \right)}^2}}}\\ = \dfrac{{2\left( {{{\cos }^2}2x + {{\sin }^2}2x} \right) + 6\cos 2x + 4\sin 2x}}{{{{\left( {\cos 2x + 3} \right)}^2}}}\\ = \dfrac{{2\left( {3\cos 2x + 2\sin 2x + 1} \right)}}{{{{\left( {\cos 2x + 3} \right)}^2}}}\end{array}$

Bước 1:

$\begin{array}{l}f'\left( x \right) = \sqrt {{x^2} - 1}  + \left( {x - 2} \right).\dfrac{x}{{\sqrt {{x^2} - 1} }}\\ = \dfrac{{\left( {{x^2} - 1} \right) + \left( {x - 2} \right).x}}{{\sqrt {{x^2} - 1} }}\\ = \dfrac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 1} }}\end{array}$

Bước 2:

$\begin{array}{l}f'\left( x \right) \le \sqrt {{x^2} - 1} \\ \Leftrightarrow \dfrac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 1} }} \le \sqrt {{x^2} - 1} \\ \Leftrightarrow \dfrac{{2{x^2} - 2x - 1}}{{\sqrt {{x^2} - 1} }} - \sqrt {{x^2} - 1}  \le 0\\ \Leftrightarrow \dfrac{{2{x^2} - 2x - 1 - \left( {{x^2} - 1} \right)}}{{\sqrt {{x^2} - 1} }} \le 0\\ \Leftrightarrow \dfrac{{{x^2} - 2x}}{{\sqrt {{x^2} - 1} }} \le 0 \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 2x \le 0\\{x^2} - 1 > 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}0 \le x \le 2\\\left[ \begin{array}{l}x > 1\\x <  - 1\end{array} \right.\end{array} \right. \Leftrightarrow 1 < x \le 2\\ =  > S = \left( {1;2} \right]\end{array}$

Bước 1:

$y' = {x^2} - mx + m$

Bước 2:

$\begin{array}{l}y' \ge 0,\forall x \in \mathbb{R}\\ \Leftrightarrow {x^2} - mx + m \ge 0\forall x \in \mathbb{R}\\ \Leftrightarrow \left\{ \begin{array}{l}1 > 0\\\Delta  \le 0\end{array} \right.\forall x \in \mathbb{R}\\ \Leftrightarrow {m^2} - 4m \le 0 \Leftrightarrow 0 \le m \le 4\end{array}$

$f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 2018} \right)$

$\begin{array}{l} \Rightarrow f'\left( x \right) = 1.\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 2018} \right) + x.1.\left( {x - 2} \right)...\left( {x - 2018} \right) + x\left( {x - 1} \right).1.\left( {x - 2} \right)...\left( {x - 2018} \right) + ... + \\x.\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 2017} \right).1\end{array}$

$\Rightarrow f'\left( 0 \right) = 1.\left( { - 1} \right)\left( { - 2} \right)...\left( { - 2018} \right) + 0 + 0 + ... + 0 = 1.2...2018 .(-1)^{2018}= 2018!$.