# 求导除法法则

## 导数除法法则

\displaystyle \left(\frac{u}{v}\right)^\prime=\frac{u^\prime v-uv^\prime}{v^2},v\not=0

\displaystyle \Delta\left(\frac{u}{v}\right)=\frac{u+\Delta u}{v+\Delta v}-\frac{u}{v}\\{}\\ =\frac{uv+v\Delta u-uv-u\Delta v}{(v+\Delta v)v}\\{}\\ =\frac{v\Delta u-u\Delta v}{(v+\Delta v)v}

\displaystyle \frac{\Delta\left(\frac{u}{v}\right)}{\Delta x}=\frac{\frac{\Delta u}{\Delta x}\cdot v-u\cdot \frac{\Delta v}{\Delta x}}{(v+\Delta v)v}\\{}\\ \because\Delta x\to0\\{}\\ \therefore\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{v}\right)=\left(\frac{\mathrm{d}u}{\mathrm{d}x}\cdot v-u\cdot \frac{\mathrm{d}v}{\mathrm{d}x}\right)\Bigg/v^2\\{}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{u}{v}\right)=\frac{u^\prime v-uv^\prime}{v^2},v\not=0

\displaystyle u=1\\{}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{v}\right)=\frac{-v^\prime}{v^2}=-v^{-2}v^\prime\\{}\\ {}\\ u=1,v=x^n\\{}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{x^n}\right)=-x^{-2n}nx^{n-1}=-nx^{-n-1}\\{}\\ \therefore\frac{\mathrm{d}}{\mathrm{d}x}x^{-n}=(-n)x^{(-n)-1}