$$$6 - \frac{a}{50}$$$的导数
您的输入
求$$$\frac{d}{da} \left(6 - \frac{a}{50}\right)$$$。
解答
和/差的导数等于导数的和/差：
$${\color{red}\left(\frac{d}{da} \left(6 - \frac{a}{50}\right)\right)} = {\color{red}\left(\frac{d}{da} \left(6\right) - \frac{d}{da} \left(\frac{a}{50}\right)\right)}$$对 $$$c = \frac{1}{50}$$$ 和 $$$f{\left(a \right)} = a$$$ 应用常数倍法则 $$$\frac{d}{da} \left(c f{\left(a \right)}\right) = c \frac{d}{da} \left(f{\left(a \right)}\right)$$$：
$$- {\color{red}\left(\frac{d}{da} \left(\frac{a}{50}\right)\right)} + \frac{d}{da} \left(6\right) = - {\color{red}\left(\frac{\frac{d}{da} \left(a\right)}{50}\right)} + \frac{d}{da} \left(6\right)$$应用幂法则 $$$\frac{d}{da} \left(a^{n}\right) = n a^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{da} \left(a\right) = 1$$$：
$$- \frac{{\color{red}\left(\frac{d}{da} \left(a\right)\right)}}{50} + \frac{d}{da} \left(6\right) = - \frac{{\color{red}\left(1\right)}}{50} + \frac{d}{da} \left(6\right)$$常数的导数是$$$0$$$:
$${\color{red}\left(\frac{d}{da} \left(6\right)\right)} - \frac{1}{50} = {\color{red}\left(0\right)} - \frac{1}{50}$$因此，$$$\frac{d}{da} \left(6 - \frac{a}{50}\right) = - \frac{1}{50}$$$。
答案
$$$\frac{d}{da} \left(6 - \frac{a}{50}\right) = - \frac{1}{50}$$$A