Cho f(x) và g(x) là hai hàm số có đạo hàm tại x₀. Xét hàm số h(x) = f(x) + g(x).

Ta có  $\frac{h\left(x\right)-h\left(x_0\right)}{x-x_0}=\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}+\frac{g\left(x\right)-g\left(x_0\right)}{x-x_0}$.

Nên  $h^{\text{'}}\left(x\right)=\underset{x\to x_0}{\lim }\frac{h\left(x\right)-h\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}+\underset{x\to x_0}{\lim }\frac{g\left(x\right)-g\left(x_0\right)}{x-x_0}=\mathrm{...}+\mathrm{...}$

Chọn biểu thức thích hợp thay cho chỗ chấm để tìm h'(x₀).

Ta có  $\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=f^{\text{'}}\left(x_0\right)$ và  $\underset{x\to x_0}{\lim }\frac{g\left(x\right)-g\left(x_0\right)}{x-x_0}=g^{\text{'}}\left(x_0\right)$ nên h'(x₀) = f'(x₀) + g'(x₀).

Do đó  $h\left(x\right)=\underset{x\to x_0}{\lim }\frac{h\left(x\right)-h\left(x_0\right)}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}+\underset{x\to x_0}{\lim }\frac{g\left(x\right)-g\left(x_0\right)}{x-x_0}=f^{\text{'}}\left(x_0\right)+g^{\text{'}}\left(x_0\right)$.