商の微分

$ \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}

$ y=\frac{f(x)}{g(x)}

$ g(x) \neq 0

$ \frac{\mathrm{d}f}{\mathrm{d}x} = \frac{\mathrm{d}g}{\mathrm{d}x}y + g\frac{\mathrm{d}y}{\mathrm{d}x}

変形して整理する:

code:tex

\begin{aligned}

\frac{\mathrm{d}y}{\mathrm{d}x}

&= \frac{1}{g}\left(\frac{\mathrm{d}f}{\mathrm{d}x} - \frac{\mathrm{d}g}{\mathrm{d}x}y\right)\\

&= \frac{1}{g}\left(\frac{\mathrm{d}f}{\mathrm{d}x} - \frac{\mathrm{d}g}{\mathrm{d}x}\frac{f}{g}\right)\\

&= \frac{1}{g^2}\left(\frac{\mathrm{d}f}{\mathrm{d}x}g - \frac{\mathrm{d}g}{\mathrm{d}x}f\right)

\end{aligned}