Cho biết  $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$. Dùng định nghĩa tính đạo hàm của hàm số y = sinx.

Có  $y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$ $=\underset{x\to x_0}{\lim }\frac{\sin x-\sin x_0}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{2cos\left(\frac{x+x_0}{2}\right)\sin \left(\frac{x-x_0}{2}\right)}{x-x_0}$

$=\underset{x\to x_0}{\lim }\left[2cos\left(\frac{x+x_0}{2}\right)\right]\cdot \underset{x\to x_0}{\lim }\left[\frac{1}{2}\cdot \frac{\sin \left(\frac{x-x_0}{2}\right)}{\frac{x-x_0}{2}}\right]$

 $=\left[2cos\left(\frac{2x_0}{2}\right)\right]\cdot \frac{1}{2}=\cos x_0$ (do  $\underset{x\to x_0}{\lim }\frac{\sin \left(\frac{x-x_0}{2}\right)}{\frac{x-x_0}{2}}=1$).

Vậy y' = (sinx)' = cosx.