数学分析 导数的四则运算法则及多个函数的推广

导数的四则运算法则

设 f(x),g(x) 在 x 可导， 则：

1) (f+g)′=f′+g′

2) (f⋅g)′=f′⋅g+f⋅g′

3) (fg)′=f′⋅g−f⋅g′g2

证明

1) (f+g)(x+Δx)−(f+g)(x)Δx

=f(x+Δx)+g(x+Δx)−[f(x)+g(x)]Δx

=[f(x+Δx)−f(x)]+[g(x+Δx)−g(x)]Δx

=f(x+Δx)−f(x)Δx+g(x+Δx)−g(x)Δx

→f′(x)+g′(x),Δx→0

2) (f⋅g)(x+Δx)−(f⋅g)(x)Δx

=f(x+Δx)g(x+Δx)−f(x)g(x)Δx

=[f(x+Δx)−f(x)]g(x+Δx)+f(x)g(x+Δx)−f(x)g(x)Δx

=[f(x+Δx)−f(x)]g(x+Δx)+f(x)[g(x+Δx)−g(x)]Δx

=[f(x+Δx)−f(x)]g(x+Δx)Δx+f(x)[g(x+Δx)−g(x)]Δx

=f(x+Δx)−f(x)Δx⋅g(x+Δx)+f(x)⋅g(x+Δx)−g(x)Δx

→f′(x)g(x)+f(x)g′(x),Δx→0

3) (fg)(x+Δx)−(fg)(x)Δx

=f(x+Δx)g(x+Δx)−f(x)g(x)Δx

=f(x+Δx)g(x)−f(x)g(x+Δx)Δx⋅g(x)⋅g(x+Δx)

=[f(x+Δx)−f(x)]g(x)−f(x)[g(x+Δx)−g(x)]Δx⋅g(x)⋅g(x+Δx)

=1g(x)⋅g(x+Δx)⋅[f(x+Δx)−f(x)]g(x)−f(x)[g(x+Δx)−g(x)]Δx

=1g(x)⋅g(x+Δx)⋅[f(x+Δx)−f(x)]g(x)Δx−f(x)[g(x+Δx)−g(x)]Δx

=1g(x)⋅g(x+Δx)⋅[f(x+Δx)−f(x)Δxg(x)−f(x)g(x+Δx)−g(x)Δx]

→f′(x)g(x)−f(x)g′(x)[g(x)]2,Δx→0

多个函数的线性组合的导函数

∀n∈N,n≥1,[∑i=1ncifi(x)]′=∑i=1ncif′i(x)

证明

n=1 时显然成立。

设 n 时成立。则 n+1时，

[∑i=1n+1cifi(x)]′=[∑i=1ncifi(x)+cn+1fn+1(x)]′

=[∑i=1ncifi(x)]′+cn+1f′n+1(x)

=∑i=1ncif′i(x)+cn+1f′n+1(x)

=∑i=1n+1cif′i(x)

多个函数的乘积的导函数

∀n∈N,n≥1,[∏i=1nfi(x)]′=∑j=1nf′j(x)⎡⎣⎢⎢⎢∏i=1i≠jnfi(x)⎤⎦⎥⎥⎥

证明

n=1 时显然成立。

设 n 时成立。则 n+1时，

[∏i=1n+1fi(x)]′=[∏i=1nfi(x)⋅fn+1(x)]′

=[∏i=1nfi(x)]′⋅fn+1(x)+∏i=1nfi(x)⋅f′n+1(x)

=∑j=1nf′j(x)⎡⎣⎢⎢⎢∏i=1i≠jnfi(x)⎤⎦⎥⎥⎥⋅fn+1(x)+∏i=1nfi(x)⋅f′n+1(x)

=∑j=1nf′j(x)⎡⎣⎢⎢⎢∏i=1i≠jn+1fi(x)⎤⎦⎥⎥⎥+f′n+1(x)∏i=1nfi(x)

=∑j=1n+1f′j(x)⎡⎣⎢⎢⎢∏i=1i≠jn+1fi(x)⎤⎦⎥⎥⎥