常数函数 f ( x ) = C f(x)=C f(x)=C， f ′ ( x ) = 0 f'(x)=0 f′(x)=0

f ( x ) = x n ( n ∈ N ∗ ) f(x)=x^n(n\in N^{*}) f(x)=xn(n∈N∗) 当 n = 1 n=1 n=1 时， f ( x ) = 1 f(x)=1 f(x)=1 当 n > 1 n>1 n>1 时， f ′ ( x ) = lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim ⁡ Δ x → 0 ( x + Δ x ) n − x n Δ x = lim ⁡ Δ x → 0 n x n − 1 + ( n 2 ) x n − 2 Δ x + ⋯ + Δ x n − 1 = n x n − 1 f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\ =\lim_{\Delta x\to 0}\frac{(x+\Delta x)^n-x^n}{\Delta x}\\ =\lim_{\Delta x\to 0}nx^{n-1}+\binom{n}{2}x^{n-2}\Delta x+\dots +\Delta x^{n-1}=nx^{n-1} f′(x)=Δx→0lim​Δxf(x+Δx)−f(x)​=Δx→0lim​Δx(x+Δx)n−xn​=Δx→0lim​nxn−1+(2n​)xn−2Δx+⋯+Δxn−1=nxn−1

幂函数 f ( x ) = x μ ( μ ∈ R ) f(x)=x^{\mu}(\mu\in R) f(x)=xμ(μ∈R)，设 x x x 在 f ( x ) f(x) f(x) 的定义域内且 x ≠ 0 x\neq 0 x​=0 引理1： lim ⁡ x → 0 l o g a ( 1 + x ) x = lim ⁡ x → 0 l o g a ( 1 + x ) 1 x = 1 ln ⁡ ( a ) \lim_{x\to 0}\frac{log_a(1+x)}{x}=\lim_{x\to 0}log_a(1+x)^{\frac{1}{x}}=\frac{1}{\ln (a)} x→0lim​xloga​(1+x)​=x→0lim​loga​(1+x)x1​=ln(a)1​

引理2： lim ⁡ x → 0 ( 1 + x ) μ − 1 x = lim ⁡ x → 0 ( 1 + x ) μ − 1 ln ⁡ ( 1 + x ) μ ∗ μ ln ⁡ ( 1 + x ) x lim ⁡ t → 0 t ln ⁡ ( 1 + t ) ∗ lim ⁡ x → 0 μ ln ⁡ ( 1 + x ) x = μ \lim_{x\to0}\frac{(1+x)^{\mu}-1}{x}\\ = \lim_{x\to 0}\frac{(1+x)^{\mu}-1}{\ln (1+x)^{\mu}}*\frac{\mu \ln (1+x)}{x}\\\lim_{t\to 0}\frac{t}{\ln (1+t)}*\lim_{x\to 0}\frac{\mu \ln (1+x)}{x}=\mu x→0lim​x(1+x)μ−1​=x→0lim​ln(1+x)μ(1+x)μ−1​∗xμln(1+x)​t→0lim​ln(1+t)t​∗x→0lim​xμln(1+x)​=μ 故 f ′ ( x ) = lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim ⁡ Δ x → 0 ( x + Δ x ) μ − x μ Δ x = lim ⁡ Δ x → 0 x μ − 1 ∗ ( 1 + Δ x x ) μ − 1 Δ x x = μ x μ − 1 f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\ =\lim_{\Delta x\to 0}\frac{(x+\Delta x)^{\mu}-x^{\mu}}{\Delta x}\\ =\lim_{\Delta x\to 0}x^{\mu -1}*\frac{(1+\frac{\Delta x}{x})^{\mu}-1}{\frac{\Delta x}{x}}=\mu x^{\mu-1} f′(x)=Δx→0lim​Δxf(x+Δx)−f(x)​=Δx→0lim​Δx(x+Δx)μ−xμ​=Δx→0lim​xμ−1∗xΔx​(1+xΔx​)μ−1​=μxμ−1

f ( x ) = sin ⁡ x f(x)=\sin x f(x)=sinx 的导数 引理 lim ⁡ x → 0 sin ⁡ x x = 1 \lim_{x\to 0}\frac{\sin x}{x}=1 limx→0​xsinx​=1 运用夹逼法，得面积关系有 tan ⁡ x > x > sin ⁡ x → cos ⁡ x < sin ⁡ x x < 1 \tan x>x>\sin x \to \cos xsinx→cosx 0 , a ≠ 1 ) f(x)=a^{x}(a>0,a\neq 1) f(x)=ax(a>0,a​=1) 的导数 引理 lim ⁡ x → 0 a x − 1 x = ln ⁡ a \lim_{x\to 0}\frac{a^x-1}{x}=\ln a limx→0​xax−1​=lna 令 t = a x − 1 t=a^x-1 t=ax−1, lim ⁡ x → 0 a x − 1 x = lim ⁡ t → 0 t l o g a ( t + 1 ) = ln ⁡ a \lim_{x\to 0}\frac{a^x-1}{x}=\lim_{t\to 0}\frac{t}{log_a(t+1)}=\ln a limx→0​xax−1​=limt→0​loga​(t+1)t​=lna f ′ ( x ) = lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = a x lim ⁡ Δ x → 0 a Δ x − 1 Δ x = ln ⁡ a ∗ a x f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\= a^x\lim_{\Delta x\to 0}\frac{a^{\Delta x}-1}{\Delta x}=\ln a*a^x f′(x)=Δx→0lim​Δxf(x+Δx)−f(x)​=axΔx→0lim​ΔxaΔx−1​=lna∗ax

f ( x ) = log ⁡ a x ( a > 0 , a ≠ 1 ) f(x)=\log_a x(a>0,a\neq 1) f(x)=loga​x(a>0,a​=1) 的导数 f ′ ( x ) = lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x = lim ⁡ Δ x → 0 1 Δ x log ⁡ a x + Δ x x = lim ⁡ Δ x → 0 1 x ∗ x Δ x log ⁡ a ( 1 + Δ x x ) = 1 x ln ⁡ a f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}\\=\lim_{\Delta x\to 0}\frac{1}{\Delta x}\log_a\frac{x+\Delta x}{x}=\\ \lim_{\Delta x\to 0}\frac{1}{x}*\frac{x}{\Delta x}\log_a(1+\frac{\Delta x}{x})=\frac{1}{x\ln a} f′(x)=Δx→0lim​Δxf(x+Δx)−f(x)​=Δx→0lim​Δx1​loga​xx+Δx​=Δx→0lim​x1​∗Δxx​loga​(1+xΔx​)=xlna1​ 于是我们有 ( ln ⁡ x ) ′ = 1 x (\ln x)'=\frac{1}{x} (lnx)′=x1​