`~rai~`
\(y=f(x)=x^2+3x\\\text{C1:Giả sử }\Delta x\text{ là số gia của đối số tại }x_0=-1.\text{Ta có:}\\\Delta y=f(-1+\Delta x)-f(-1)\\\quad\quad=(-1+\Delta x)^2+3(-1+\Delta x)+2\\\quad\quad=1-2\Delta x+(\Delta x)^2-3+3\Delta x+2\\\quad\quad=\Delta x(\Delta x+1).\\\dfrac{\Delta y}{\Delta x}=\dfrac{\Delta x(\Delta x+1)}{\Delta x}=\Delta x+1.\\\Rightarrow \lim\limits_{\Delta x\to 0}\dfrac{\Delta y}{\Delta x}=\lim\limits_{\Delta x\to 0} \Delta x+1=1.\\\text{Vậy }f'(-1)=1.\\\text{Cách 2:Áp dụng công thức:}\\f'(x_0)=\lim\limits_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}\\\Rightarrow f(-1)=\lim\limits_{x\to -1}\dfrac{x^2+3x+2}{x+1}\\\quad\quad\quad\quad=\lim\limits_{x\to -1}\dfrac{(x+1)(x+2)}{x+1}\\\quad\quad\quad\quad=\lim\limits_{x\to -1}x+2\\\quad\quad\quad\quad=1.\\\text{Vậy }f'(-1)=1.\)
