参考にしたサイト➔ここだよ^^
解説:
\( \hspace{35px} \begin{eqnarray} \biggl( \dfrac{1}{\tan x} \biggr)' & = & \dfrac{0 - (\tan x)'}{\tan^2 x} \\ & = & -\dfrac{1}{\sin^2 x} \end{eqnarray} \)解説:
\( \hspace{35px} \begin{eqnarray} y & = & x^x \\ \log y & = & x\log x \\ \dfrac{y'}{y} & = & \log x + 1 \\ y' & = & y(\log x + 1) \\ & = & x^x(\log x + 1) \end{eqnarray} \)※別解:
\( \hspace{35px} \begin{eqnarray} (x^x)' & = & (e^{x\log x})' \\ & = & e^{x\log x}(x\log x)' \\ & = & x^x(\log x + 1) \\ \end{eqnarray} \)解説:
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{4} \sqrt{x} dx & = & \left[ \dfrac{2}{3}x^{\frac{3}{2}} \right]_{0}^{4} \\ & = & \dfrac{16}{3} \\ \end{eqnarray} \)解説:
\( \hspace{35px} \begin{eqnarray} \displaystyle \int_{0}^{\frac{1}{2}} \dfrac{1}{1 - x^2} dx & = & \displaystyle \int_{0}^{\frac{1}{2}} \dfrac{1}{2}\biggr( \dfrac{1}{1 + x} + \dfrac{1}{1 - x} \biggl) dx \\ & = & \dfrac{1}{2}\biggl[ \log|1 + x| - \log|1 - x| \biggr]_{0}^{\frac{1}{2}} \\ & = & \dfrac{1}{2}\biggl[ \log\biggl| \dfrac{1 + x}{1 - x} \biggr| \biggr]_{0}^{\frac{1}{2}} \\ & = & \dfrac{1}{2}\log 3 \end{eqnarray} \)解説:
\( \displaystyle \int e^x\sin x\cos x dx = I \) とすると