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$$$\sec^{3}{\left(u \right)}$$$の導関数
入力内容
$$$\frac{d}{du} \left(\sec^{3}{\left(u \right)}\right)$$$ を求めよ。
解答
関数$$$\sec^{3}{\left(u \right)}$$$は、2つの関数$$$f{\left(v \right)} = v^{3}$$$と$$$g{\left(u \right)} = \sec{\left(u \right)}$$$の合成$$$f{\left(g{\left(u \right)} \right)}$$$である。
連鎖律 $$$\frac{d}{du} \left(f{\left(g{\left(u \right)} \right)}\right) = \frac{d}{dv} \left(f{\left(v \right)}\right) \frac{d}{du} \left(g{\left(u \right)}\right)$$$ を適用する:
$${\color{red}\left(\frac{d}{du} \left(\sec^{3}{\left(u \right)}\right)\right)} = {\color{red}\left(\frac{d}{dv} \left(v^{3}\right) \frac{d}{du} \left(\sec{\left(u \right)}\right)\right)}$$冪法則 $$$\frac{d}{dv} \left(v^{n}\right) = n v^{n - 1}$$$ を $$$n = 3$$$ に対して適用する:
$${\color{red}\left(\frac{d}{dv} \left(v^{3}\right)\right)} \frac{d}{du} \left(\sec{\left(u \right)}\right) = {\color{red}\left(3 v^{2}\right)} \frac{d}{du} \left(\sec{\left(u \right)}\right)$$元の変数に戻す:
$$3 {\color{red}\left(v\right)}^{2} \frac{d}{du} \left(\sec{\left(u \right)}\right) = 3 {\color{red}\left(\sec{\left(u \right)}\right)}^{2} \frac{d}{du} \left(\sec{\left(u \right)}\right)$$正割関数の導関数は$$$\frac{d}{du} \left(\sec{\left(u \right)}\right) = \tan{\left(u \right)} \sec{\left(u \right)}$$$:
$$3 \sec^{2}{\left(u \right)} {\color{red}\left(\frac{d}{du} \left(\sec{\left(u \right)}\right)\right)} = 3 \sec^{2}{\left(u \right)} {\color{red}\left(\tan{\left(u \right)} \sec{\left(u \right)}\right)}$$したがって、$$$\frac{d}{du} \left(\sec^{3}{\left(u \right)}\right) = 3 \tan{\left(u \right)} \sec^{3}{\left(u \right)}$$$。
解答
$$$\frac{d}{du} \left(\sec^{3}{\left(u \right)}\right) = 3 \tan{\left(u \right)} \sec^{3}{\left(u \right)}$$$A