要找到 cos ( x ) {\displaystyle \cos(x)} 的导数。
正如在证明 正弦函数的导数 中，第一项的极限为 sin ( x ) {\displaystyle \sin(x)} ，第二项的极限为 1。因此
因此
等等，无穷无尽。
![{\displaystyle {\frac {d}{dx}}\cos(x)=\lim _{h\to 0}{\frac {\cos(x+h)-\cos(x)}{h}}=-\lim _{h\to 0}{\frac {2\sin \left(x+{\frac {h}{2}}\right)\sin \left({\frac {h}{2}}\right)}{h}}=-\lim _{h\to 0}\left[\sin \left(x+{\frac {h}{2}}\right){\frac {\sin \left({\frac {h}{2}}\right)}{\frac {h}{2}}}\right]=-\lim _{h\to 0}\left[\sin \left(x+{\frac {h}{2}}\right)\right]\cdot \lim _{h\to 0}\left[{\frac {\sin \left({\frac {h}{2}}\right)}{\frac {h}{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6290f7b65de7fb5fd71271d3365c2b63c6e75642)
![{\displaystyle {\frac {d}{dx}}{\big [}\cos(x){\big ]}=-\sin(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee6354b64afce92f2b1cadb5a495f5e1c88ceb25)
![{\displaystyle {\frac {d}{dx}}{\big [}\sin(x){\big ]}=\cos(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e0926652b30841b3f9411fa26d6ff94ef0695d9)
![{\displaystyle {\frac {d}{dx}}{\big [}-\sin(x){\big ]}=-\cos(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0696bd65cd43b2ea95dcaca878ef065c88f333a)
![{\displaystyle {\frac {d}{dx}}{\big [}-\cos(x){\big ]}=\sin(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc077e992d87ea1d588af0562c2e7e46cecddfcd)