cot(ax)の微分

\(\cot{\left(ax\right)}\)の微分

$$\begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d}x}\cot{\left(ax\right)} &=&\frac{\mathrm{d}}{\mathrm{d}x}\left\{\tan{\left(ax\right)}\right\}^{-1}\;\ldots\;a,x\in\mathbb{R} \\&=&\frac{\mathrm{d}}{\mathrm{d}u}\left\{\tan{\left(u\right)}\right\}^{-1}\frac{\mathrm{d}u}{\mathrm{d}x} \;\ldots\;u=ax,\;\frac{\mathrm{d}u}{\mathrm{d}x}=a \\&=&\frac{\mathrm{d}}{\mathrm{d}v}\left\{ v\left(u\right)\right\}^{-1}\frac{\mathrm{d}v}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} \;\ldots\;v=\tan{\left(u\right)},\;\frac{\mathrm{d}v}{\mathrm{d}u}=1+\tan^2{\left(u\right)} \\&=&-\left\{ v\left(u\right)\right\}^{-2}\frac{\mathrm{d}v}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&-\frac{1}{\tan^2{\left(u\right)}}\cdot \left(1+\tan^2{\left(u\right)}\right)\frac{\mathrm{d}u}{\mathrm{d}x} \\&=&-\frac{1}{\tan^2{\left(ax\right)}}\cdot \left(1+\tan^2{\left(ax\right)}\right)\cdot a \\&=&-\frac{a\left(1+\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}\right)}{\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}} \\&=&-\frac{\cos^2{\left(ax\right)}a\left(1+\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}\right)}{\sin^2{\left(ax\right)}} \\&=&-\frac{a\left(\cos^2{\left(ax\right)}+\cos^2{\left(ax\right)}\frac{\sin^2{\left(ax\right)}}{\cos^2{\left(ax\right)}}\right)}{\sin^2{\left(ax\right)}} \\&=&-\frac{a\left(\cos^2{\left(ax\right)}+\sin^2{\left(ax\right)}\right)}{\sin^2{\left(ax\right)}} \\&=&-\frac{a\cdot 1}{\sin^2{\left(ax\right)}} \\&=&-\frac{a}{\sin^2{\left(ax\right)}} \end{eqnarray}$$