|
|
|
|
|
|
|
|
|
|
|
|
$$$\frac{1}{\sin{\left(x \right)}}$$$的导数
您的输入
求$$$\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right)$$$。
解答
函数$$$\frac{1}{\sin{\left(x \right)}}$$$是两个函数$$$f{\left(u \right)} = \frac{1}{u}$$$和$$$g{\left(x \right)} = \sin{\left(x \right)}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。
应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：
$${\color{red}\left(\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right) \frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}$$应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = -1$$$:
$${\color{red}\left(\frac{d}{du} \left(\frac{1}{u}\right)\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right) = {\color{red}\left(- \frac{1}{u^{2}}\right)} \frac{d}{dx} \left(\sin{\left(x \right)}\right)$$返回到原变量：
$$- \frac{\frac{d}{dx} \left(\sin{\left(x \right)}\right)}{{\color{red}\left(u\right)}^{2}} = - \frac{\frac{d}{dx} \left(\sin{\left(x \right)}\right)}{{\color{red}\left(\sin{\left(x \right)}\right)}^{2}}$$正弦函数的导数为 $$$\frac{d}{dx} \left(\sin{\left(x \right)}\right) = \cos{\left(x \right)}$$$:
$$- \frac{{\color{red}\left(\frac{d}{dx} \left(\sin{\left(x \right)}\right)\right)}}{\sin^{2}{\left(x \right)}} = - \frac{{\color{red}\left(\cos{\left(x \right)}\right)}}{\sin^{2}{\left(x \right)}}$$因此，$$$\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right) = - \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$。
答案
$$$\frac{d}{dx} \left(\frac{1}{\sin{\left(x \right)}}\right) = - \frac{\cos{\left(x \right)}}{\sin^{2}{\left(x \right)}}$$$A