我们都知道， e x e^x ex的导数为 e x e^x ex，但怎么证明呢？

设 y = e x y=e^x y=ex

则 y ′ = lim ⁡ Δ x → 0 e x + Δ x − e x Δ x y'=\lim\limits_{\Delta x\rightarrow 0}\dfrac{e^{x+\Delta x}-e^x}{\Delta x} y′=Δx→0lim​Δxex+Δx−ex​

= e x × lim ⁡ Δ x → 0 e Δ x − 1 Δ x \qquad =e^x\times\lim\limits_{\Delta x\rightarrow 0}\dfrac{e^{\Delta x}-1}{\Delta x} =ex×Δx→0lim​ΔxeΔx−1​

∵ e = lim ⁡ x → 0 ( 1 + x ) 1 x \quad\because e=\lim\limits_{x\rightarrow 0} (1+x)^{\frac 1x} ∵e=x→0lim​(1+x)x1​

∴ lim ⁡ Δ x → 0 ( 1 + Δ x ) 1 Δ x = e \quad\therefore \lim\limits_{\Delta x\rightarrow 0}(1+\Delta x)^{\frac{1}{\Delta x}}=e ∴Δx→0lim​(1+Δx)Δx1​=e

y ′ = e x × lim ⁡ Δ x → 0 ( ( 1 + Δ x ) 1 Δ x ) Δ x − 1 Δ x \quad y'=e^x\times \lim\limits_{\Delta x\rightarrow 0}\dfrac{((1+\Delta x)^{\frac{1}{\Delta x}})^{\Delta x}-1}{\Delta x} y′=ex×Δx→0lim​Δx((1+Δx)Δx1​)Δx−1​

= e x × lim ⁡ Δ x → 0 ( 1 + Δ x ) − 1 Δ x \qquad =e^x\times \lim\limits_{\Delta x\rightarrow 0}\dfrac{(1+\Delta x)-1}{\Delta x} =ex×Δx→0lim​Δx(1+Δx)−1​

= e x × lim ⁡ Δ x → 0 Δ x Δ x \qquad =e^x\times \lim\limits_{\Delta x\rightarrow 0}\dfrac{\Delta x}{\Delta x} =ex×Δx→0lim​ΔxΔx​

= e x \qquad =e^x =ex

所以 e x e^x ex的导数为 e x e^x ex

证明： ( a x ) ′ = a x ln ⁡ a (a^x)'=a^x\ln a (ax)′=axlna

解： 已证出 ( e x ) ′ = e x (e^x)'=e^x (ex)′=ex，则 a x = ( e ln ⁡ a ) x = e x ln ⁡ a a^x=(e^{\ln a})^x=e^{x\ln a} ax=(elna)x=exlna

所以 ( a x ) ′ = ( e x ln ⁡ a ) ′ = e x ln ⁡ a × ln ⁡ a = a x ln ⁡ a (a^x)'=(e^{x\ln a})'=e^{x \ln a}\times \ln a=a^x\ln a (ax)′=(exlna)′=exlna×lna=axlna