Khái niệm đạo hàm

$\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{x^2}}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} x = 0\\\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{ - {x^2}}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \left( { - x} \right) = 0\end{array} \right.$ $\Rightarrow \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = 0$

$\Rightarrow$ Hàm số $y = f\left( x \right)$ có đạo hàm tại $x = 0.$

$\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{g\left( x \right) - g\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt x }}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{\sqrt x }}= + \infty$

$\Rightarrow$ Hàm số $y = g\left( x \right)$ không có đạo hàm tại $x = 0.$

Ta có:

$\begin{array}{l}f'(0)=\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{4{x^2} + 8}} - \sqrt {8{x^2} + 4} }}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt[3]{{4{x^2} + 8}} - 2}}{{{x^2}}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {8{x^2} + 4}  - 2}}{{{x^2}}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{{4{x^2}}}{{{x^2}\left( {{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4} \right)}} - \mathop {\lim }\limits_{x \to 0} \dfrac{{8{x^2}}}{{{x^2}\left( {\sqrt {8{x^2} + 4}  + 2} \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \dfrac{4}{{{{\sqrt[3]{{4{x^2} + 8}}}^2} + 2\sqrt[3]{{4{x^2} + 8}} + 4}} - \mathop {\lim }\limits_{x \to 0} \dfrac{8}{{\sqrt {8{x^2} + 4}  + 2}} = \dfrac{1}{3} - 2 =  - \dfrac{5}{3}\end{array}$

Dựa vào đồ thị hàm số ta thấy $\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = 1,\,\,\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = 0 \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) \Rightarrow$ Không tồn tại $\mathop {\lim }\limits_{x \to 1} f\left( x \right)$, hàm số không liên tục tại $x = 1.$

Ngoài ra tại các điểm $x=0,x=2,x=3$ thì hàm số đều có đạo hàm.

Vậy hàm số không có đạo hàm tại $x = 1.$

Ta có: $\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{\dfrac{{{x^3} - 4{x^2} + 3x}}{{{x^2} - 3x + 2}} - 0}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{x\left( {x - 3} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x - 2} \right)}}$ $= \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} =  + \infty$

$\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{\dfrac{{{x^3} - 4{x^2} + 3x}}{{{x^2} - 3x + 2}} - 0}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{x\left( {x - 3} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}\left( {x - 2} \right)}}$ $= \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}} =  - \infty$

Do đó không tồn tại giới hạn $\mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}$

Vậy hàm số không có đạo hàm tại $x = 1$.

$f'\left( { - 1} \right) = \mathop {\lim }\limits_{x \to \left( { - 1} \right)} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}}$

Ta có:

$\begin{array}{l}\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \dfrac{{\dfrac{{{x^2} + x + 1}}{x} + 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \dfrac{{{x^2} + 2x + 1}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \dfrac{{x + 1}}{x} = 0\\\mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{\dfrac{{{x^2} - x - 1}}{x} + 1}}{{x + 1}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{{x^2} - 1}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{x - 1}}{x} = 2\\ \Rightarrow \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ + }} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}} \ne \mathop {\lim }\limits_{x \to {{\left( { - 1} \right)}^ - }} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}}\end{array}$

Do đó không tồn tại  $\mathop {\lim }\limits_{x \to \left( { - 1} \right)} \dfrac{{f\left( x \right) - f\left( { - 1} \right)}}{{x + 1}}$, vậy không tồn tại đạo hàm của hàm số tại ${x_0} =  - 1$.

Ta có: $y = \dfrac{{\left| x \right|}}{{x + 1}} = \left\{ \begin{array}{l}\dfrac{x}{{x + 1}}\,\,\,khi\,x \ge 0\\\dfrac{{ - x}}{{x + 1}}\,\,\,khi\,\,x < 0\end{array} \right.$

Ta có $\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{x}{{x + 1}} = 0 = f\left( 0 \right)\\\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{ - x}}{{x + 1}} = 0\end{array} \right. \Rightarrow \mathop {\lim }\limits_{x \to x_0^ + } f\left( x \right) = \mathop {\lim }\limits_{x \to x_0^ - } f\left( x \right) = f\left( 0 \right) = 0 \Rightarrow$ Hàm số liên tục tại $x = 0.$

$f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}$ 

Ta có:

$\left\{ \begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\dfrac{x}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{1}{{x + 1}} = 1\\\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\dfrac{{ - x}}{{x + 1}} - 0}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{ - 1}}{{x + 1}} =  - 1\end{array} \right. \Rightarrow x = {x_0} \Rightarrow \mathop {\lim }\limits_{x \to x_0^ + } \dfrac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} \ne \mathop {\lim }\limits_{x \to x_0^ - } \dfrac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} \Rightarrow$ Hàm số không tồn tại đạo hàm tại $x = 0.$

Để hàm số có đạo hàm của hàm số tại điểm $x = 1$ thì trước hết hàm số phải liên tục tại $x = 1,$ tức là $\mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right) \Leftrightarrow \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} - 1}}{{x - 1}} = a$ $\Leftrightarrow \mathop {\lim }\limits_{x \to 1} \left( {x + 1} \right) = a \Leftrightarrow 2 = a$

Khi đó hàm số có dạng: $f\left( x \right) = \left\{ \begin{array}{l}\dfrac{{{x^2} - 1}}{{x - 1}}\,\,khi\,\,x \ne 1\\2\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,x = 1\end{array} \right.$

$\Rightarrow f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to 1} \dfrac{{\dfrac{{{x^2} - 1}}{{x - 1}} - 2}}{{x - 1}}$ $= \mathop {\lim }\limits_{x \to 1} \dfrac{{x + 1 - 2}}{{x - 1}} = 1$

Vậy $a = 2.$

Trước tiên hàm số phải liên tục tại $x = 0.$

Ta có:

$\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{x^2} + 1}}{{x + 1}} = 1= f\left( 0 \right)\\\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {ax + b} \right) = b \end{array}$

Để hàm số liên tục tại $x = 0$ thì $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right) \Leftrightarrow b = 1$

Khi đó ta có $f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{x}$

Ta có

$\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\dfrac{{{x^2} + 1}}{{x + 1}} - 1}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{{x^2} - x}}{{x\left( {x + 1} \right)}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{x - 1}}{{x + 1}} =  - 1\\\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left( {ax + 1} \right) - 1}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} a = a\end{array}$

Để hàm số có đạo hàm tại $x = 0$ thì  $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{x} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{x} \Leftrightarrow a =  - 1$

Vậy $a =  - 1,b = 1$.

$\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a{x^2} + bx} \right) = a + b = f\left( 1 \right)\\\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {2x - 1} \right) = 1\end{array}$

Để hàm số liên tục tại $x = 1$ thì $\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = f\left( 1 \right) \Leftrightarrow a + b = 1\,\,\,\left( 1 \right)$

Khi đó ta có: $f'\left( 1 \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}$

$\begin{array}{l}\mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{a{x^2} + bx - \left( {a + b} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{a\left( {{x^2} - 1} \right) + b\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \left[ {a\left( {x + 1} \right) + b} \right] = 2a + b\\\mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{2x - 1 - \left( {a + b} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{2x - 2}}{{x - 1}} = 2\end{array}$

Để hàm số có đạo hàm tại $x = 1$ thì $f'\left( 1 \right) = \mathop {\lim }\limits_{x \to {1^ + }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to {1^ - }} \dfrac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} \Leftrightarrow 2a + b = 2\,\,\,\left( 2 \right)$

Từ (1) và (2) ta có hệ: $\left\{ \begin{array}{l}a + b = 1\\2a + b = 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b = 0\end{array} \right.$

Một hàm số liên tục tại $x_0$ chưa chắc có đạo hàm tại điểm đó, hơn nữa

$\mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{x\sin \dfrac{\pi }{x} - 0}}{x}$ $= \mathop {\lim }\limits_{x \to 0} \sin \dfrac{\pi }{x} =  + \infty$ $\Rightarrow$ hàm số không có đạo hàm tại $x = 0.$

Lập luận trên sai từ bước 4.

$\begin{array}{l}f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{x\left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 1000} \right) - 0}}{x}\\ = \mathop {\lim }\limits_{x \to 0} \left( {x - 1} \right)\left( {x - 2} \right)...\left( {x - 1000} \right) = \mathop {\lim }\limits_{x \to 0} \left( { - 1} \right)\left( { - 2} \right)\left( { - 3} \right)...\left( { - 1000} \right) = {\left( { - 1} \right)^{1000}}.1000! = 1000!\end{array}$

Để hàm số có đạo hàm tại $x = 1$ thì trước hết hàm số phải liên tục tại $x = 1.$

Ta có: $f\left( 0 \right) = 1$

$\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {a{x^2} + bx + 1} \right) = 1 = f\left( 0 \right)\\\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {a\sin x + b\cos x} \right) = b\end{array}$

Để hàm số liên tục tại $x = 1$ thì $\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = f\left( 0 \right) \Leftrightarrow b = 1$

Khi đó ta có: $f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}}$

$\begin{array}{l}\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{a{x^2} + x + 1 - 1}}{x} = \mathop {\lim }\limits_{x \to {0^ + }} \left( {ax + 1} \right) = 1\\\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{a\sin x + \cos x - 1}}{x} \\ = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{2a\sin \dfrac{x}{2}\cos \dfrac{x}{2} - 2{{\sin }^2}\dfrac{x}{2}}}{x} \\ = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}\mathop {\lim }\limits_{x \to {0^ - }} \left( {a\cos \dfrac{x}{2} - 2\sin \dfrac{x}{2}} \right) = a\end{array}$

Để tồn tại $f'\left( 0 \right)$ $\Leftrightarrow \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} \Leftrightarrow a = 1.$

Bước 1:

Ta có: $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x - \sin 3x}}{x} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} - \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin 3x}}{x}$

Bước 2:

$= \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} - \mathop {\lim }\limits_{x \to 0} \dfrac{{3.\sin 3x}}{{3x}} = 1 - 3 =  - 2$.