##### 导数运算 - 导数运算 - **导数运算**是对[[导数]]进行的[[运算]]和[[运算律]], 主要是与代数运算的可交换性和一些计算定理, 有[[高阶导数]], [[初等函数微积分|初等函数导数表]]等 - 和差 - $\displaystyle[f(x)\pm g(x)]'=f'(x)\pm g'(x)$ - $\displaystyle\frac{\text{d}(f\pm g)}{\text{d}x}=\frac{\text{d}f}{\text{d}x}\pm\frac{\text{d}g}{\text{d}x}$ - ${\rm d}(u\pm v)={\rm d}u\pm{\rm d}v$ - 乘积 - $\displaystyle[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)$ - $\displaystyle\frac{\text{d}(fg)}{\text{d}x}=g\frac{\text{d}f}{\text{d}x}+f\frac{\text{d}g}{\text{d}x}$ - ${\rm d}(uv)=v{\rm d}u+u{\rm d}v$ - 除法 - $\displaystyle[\frac{f(x)}{g(x)}]'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$ - $\displaystyle\frac{\text{d}(\frac{f}{g})}{\text{d}x}=\frac{g\frac{\text{d}f}{\text{d}x}-f\frac{\text{d}g}{\text{d}x}}{g^2}$ - $\displaystyle{\rm d}(\frac{u}{v})=\frac{v{\rm d}u-u{\rm d}v}{v^2}$ - 复合函数链式求导 - $\displaystyle f'[g(x)]=f'[g(x)]g'(x)$ - $\displaystyle \frac{{\rm d}f[g(x)]}{{\rm d}x}=\frac{{\rm d}f(g)}{{\rm d}g}\frac{{\rm d}g}{{\rm d}x}$ - ${\rm d}(f(u))=f'(u)g'(x){\rm d}x=f'(u){\rm d}u$ - 反函数求导 - $\displaystyle \frac{{\rm d}y}{{\rm d}x}=\frac{1}{f'(y)}$ - 求导: $\displaystyle x=f(y)$ - $\displaystyle \frac{{\rm d}}{{\rm d}x}(x)=\frac{{\rm d}}{{\rm d}x}(f(y))$ - $\displaystyle 1=\frac{{\rm d}u}{{\rm d}y}\frac{{\rm d}y}{{\rm d}x}=f'(y)\frac{{\rm d}y}{{\rm d}x}$ - $\displaystyle \frac{{\rm d}y}{{\rm d}x}=\frac{1}{f'(y)}$ - 取对数求导 - $\displaystyle y=f(x)^{g(x)}$ - $\ln(y)=g(x)\ln(f(x))$ - $\displaystyle\frac{1}{y}\frac{{\rm d}y}{{\rm d}x}=\frac{{\rm d}}{{\rm d}x}...$ - 隐函数求导 - 对方程两边一切同时求导并化简 >[!example]- 导数运算 >- 复合函数求导 > - $\displaystyle f(x)=\ln(2x+5)$ > - $\displaystyle g(x)=2x+5$ > - $\displaystyle f'[g(x)]=f'[g(x)]g'(x)=\frac{2}{2x+5}$ > - $\displaystyle f(x)=(x^2+1)^{99}$ > - $\displaystyle y=u^{99},u=x^2+1$ > - $\displaystyle\frac{{\rm d}y}{{\rm d}x}=\frac{{\rm d}y}{{\rm d}u}\frac{{\rm d}u}{{\rm d}x}=99u^{98}\cdot 2x=198(x^2+1)^{98}$ >- 隐函数求导 > - $y=\cos(x+y)$ > - $\displaystyle \frac{{\rm d}}{{\rm d}x}y=\frac{{\rm d}}{{\rm d}x}\cos(x+y)$ > - 链式求导 $\displaystyle \frac{{\rm d}y}{{\rm d}x}=-(1+\frac{{\rm d}y}{{\rm d}x})\sin(x+y)$ > - $\displaystyle \frac{{\rm d}y}{{\rm d}x}=\frac{-\sin(x+y)}{1+\sin(x+y)}$ >- 反函数求导 > - $f(x)=x^3$ > - 反函数 $y=f^{-1}(x)=x^{\frac{1}{3}}$ > - $\displaystyle \frac{{\rm d}y}{{\rm d}x}=\frac{1}{f'(y)}=\frac{1}{3y^2} =\frac{1}{3(x^{\frac{1}{3}})^2}=\frac{1}{3x^{\frac{2}{3}}}$