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$$$2 - \frac{1}{t^{2}}$$$ 的導數
您的輸入
求$$$\frac{d}{dt} \left(2 - \frac{1}{t^{2}}\right)$$$。
解答
和/差的導數等於導數的和/差：
$${\color{red}\left(\frac{d}{dt} \left(2 - \frac{1}{t^{2}}\right)\right)} = {\color{red}\left(\frac{d}{dt} \left(2\right) - \frac{d}{dt} \left(\frac{1}{t^{2}}\right)\right)}$$套用冪次法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$，取 $$$n = -2$$$：
$$- {\color{red}\left(\frac{d}{dt} \left(\frac{1}{t^{2}}\right)\right)} + \frac{d}{dt} \left(2\right) = - {\color{red}\left(- \frac{2}{t^{3}}\right)} + \frac{d}{dt} \left(2\right)$$常數的導數為$$$0$$$：
$${\color{red}\left(\frac{d}{dt} \left(2\right)\right)} + \frac{2}{t^{3}} = {\color{red}\left(0\right)} + \frac{2}{t^{3}}$$因此，$$$\frac{d}{dt} \left(2 - \frac{1}{t^{2}}\right) = \frac{2}{t^{3}}$$$。
答案
$$$\frac{d}{dt} \left(2 - \frac{1}{t^{2}}\right) = \frac{2}{t^{3}}$$$A
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