我们都知道, x n x^n xn的导数为 n x n − 1 nx^{n-1} nxn−1,那怎么证明呢?
( x n ) ′ = lim Δ x → 0 ( x + Δ x ) n − x n Δ x = x n − 1 lim Δ x → 0 ( 1 + Δ x x ) n − 1 Δ x x (x^n)'=\lim\limits_{\Delta x\rightarrow 0}\dfrac{(x+\Delta x)^n-x^n}{\Delta x}=x^{n-1}\lim\limits_{\Delta x\rightarrow 0}\dfrac{(1+\frac{\Delta x}{x})^n-1}{\frac{\Delta x}{x}} (xn)′=Δx→0limΔx(x+Δx)n−xn=xn−1Δx→0limxΔx(1+xΔx)n−1
根据无穷小替换, x → 0 x\rightarrow 0 x→0时, ( 1 + x ) a − 1 ∼ a x (1+x)^a-1\sim ax (1+x)a−1∼ax
所以 x n − 1 lim Δ x → 0 ( 1 + Δ x x ) n − 1 Δ x x = x n − 1 lim Δ x → 0 n ⋅ Δ x x Δ x x = n x n − 1 x^{n-1}\lim\limits_{\Delta x\rightarrow 0}\dfrac{(1+\frac{\Delta x}{x})^n-1}{\frac{\Delta x}{x}}=x^{n-1}\lim\limits_{\Delta x\rightarrow 0}\dfrac{n\cdot\frac{\Delta x}{x}}{\frac{\Delta x}{x}}=nx^{n-1} xn−1Δx→0limxΔx(1+xΔx)n−1=xn−1Δx→0limxΔxn⋅xΔx=nxn−1
所以 x n x^n xn的导数为 n x n − 1 nx^{n-1} nxn−1