本文主要推导了人教版A版数学选择性必修第二册上直接给出的基本的导数公式

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设 f ( x ) f(x) f(x) ， g ( x ) g(x) g(x) 均为可导函数

证明： [ f ( x ) ± g ( x ) ] ′ [f(x) \pm g(x)]' [f(x)±g(x)]′

= lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) ± [ g ( x + Δ x ) − g ( x ) ] Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x) \pm\left[g(x+\Delta x)-g(x)\right]}{\Delta x}} =Δx→0lim​Δxf(x+Δx)−f(x)±[g(x+Δx)−g(x)]​

= lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x ± lim ⁡ Δ x → 0 g ( x + Δ x ) − g ( x ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x)}{\Delta x}} \pm \lim\limits_{\Delta x\to 0}{\dfrac{g(x+\Delta x)-g(x)}{\Delta x}} =Δx→0lim​Δxf(x+Δx)−f(x)​±Δx→0lim​Δxg(x+Δx)−g(x)​

= f ′ ( x ) ± g ′ ( x ) =f'(x)\pm g'(x) =f′(x)±g′(x)

证明： [ f ( x ) g ( x ) ] ′ [f(x)g(x)]' [f(x)g(x)]′

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x =\lim\limits_{\Delta x \to 0}{\dfrac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}} =Δx→0lim​Δxf(x+Δx)g(x+Δx)−f(x)g(x)​

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x + Δ x ) + f ( x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x =\lim\limits_{\Delta x \to 0}{\dfrac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x+\Delta x)+f(x)g(x+\Delta x)-f(x)g(x)}{\Delta x}} =Δx→0lim​Δxf(x+Δx)g(x+Δx)−f(x)g(x+Δx)+f(x)g(x+Δx)−f(x)g(x)​

= lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x lim ⁡ Δ x → 0 g ( x + Δ x ) + f ( x ) lim ⁡ Δ x → 0 g ( x + Δ x ) − g ( x ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}\lim\limits_{\Delta x\to 0}{g(x+\Delta x)}+f(x)\lim\limits_{\Delta x\to 0}{\dfrac{g(x+\Delta x)-g(x)}{\Delta x}} =Δx→0lim​Δxf(x+Δx)−f(x)​Δx→0lim​g(x+Δx)+f(x)Δx→0lim​Δxg(x+Δx)−g(x)​

= f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) =f'(x)g(x)+f(x)g'(x) =f′(x)g(x)+f(x)g′(x)

证明： [ f ( x ) g ( x ) ] ′ \left[\dfrac{f(x)}{g(x)}\right]' [g(x)f(x)​]′

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)}}{\Delta x}} =Δx→0lim​Δxg(x+Δx)f(x+Δx)​−g(x)f(x)​​

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x ) − g ( x + Δ x ) f ( x ) Δ x g ( x + Δ x ) g ( x ) =\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)g(x)-g(x+\Delta x)f(x)}{\Delta xg(x+\Delta x)g(x)}} =Δx→0lim​Δxg(x+Δx)g(x)f(x+Δx)g(x)−g(x+Δx)f(x)​

= lim ⁡ Δ x → 0 f ( x + Δ x ) g ( x ) − f ( x ) g ( x ) Δ x g ( x + Δ x ) g ( x ) − lim ⁡ Δ x → 0 g ( x + Δ x ) f ( x ) − f ( x ) g ( x ) Δ x g ( x + Δ x ) g ( x ) =\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)g(x)-f(x)g(x)}{\Delta xg(x+\Delta x)g(x)}}-\lim\limits_{\Delta x \to 0}{\dfrac{g(x+\Delta x)f(x)-f(x)g(x)}{\Delta xg(x+\Delta x)g(x)}} =Δx→0lim​Δxg(x+Δx)g(x)f(x+Δx)g(x)−f(x)g(x)​−Δx→0lim​Δxg(x+Δx)g(x)g(x+Δx)f(x)−f(x)g(x)​

= lim ⁡ Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x lim ⁡ Δ x → 0 g ( x ) g ( x + Δ x ) g ( x ) − lim ⁡ Δ x → 0 g ( x + Δ x ) − g ( x ) Δ x lim ⁡ Δ x → 0 f ( x ) g ( x + Δ x ) g ( x ) =\lim\limits_{\Delta x\to 0}{\dfrac{f(x+\Delta x)-f(x)}{\Delta x}}\lim\limits_{\Delta x\to 0}{\dfrac{g(x)}{g(x+\Delta x)g(x)}} - \lim\limits_{\Delta x\to 0}{\dfrac{g(x+\Delta x)-g(x)}{\Delta x}}\lim\limits_{\Delta x\to 0}{\dfrac{f(x)}{g(x+\Delta x)g(x)}} =Δx→0lim​Δxf(x+Δx)−f(x)​Δx→0lim​g(x+Δx)g(x)g(x)​−Δx→0lim​Δxg(x+Δx)−g(x)​Δx→0lim​g(x+Δx)g(x)f(x)​

= f ′ ( x ) g ( x ) [ g ( x ) ] 2 − g ′ ( x ) f ( x ) [ g ( x ) ] 2 =f'(x)\dfrac{g(x)}{\left[g(x)\right]^2} - g'(x)\dfrac{f(x)}{\left[g(x)\right]^2} =f′(x)[g(x)]2g(x)​−g′(x)[g(x)]2f(x)​

= f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 =\dfrac{f'(x)g(x)-f(x)g'(x)}{\left[g(x)\right]^2} =[g(x)]2f′(x)g(x)−f(x)g′(x)​

证明： f ′ ( x ) f'(x) f′(x)

= lim ⁡ Δ x → 0 ( x + Δ x ) a − ( x ) a Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{(x+\Delta x)^{a}-(x)^{a}}{\Delta x}} =Δx→0lim​Δx(x+Δx)a−(x)a​

= lim ⁡ Δ x → 0 ∑ r = 0 a C a r x a − r Δ x r − C a 0 x a Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\sum\limits_{r=0}^{a}{C_{a}^{r}x^{a-r}\Delta x^{r}}-C_{a}^{0}x^{a}}{\Delta x}} =Δx→0lim​Δxr=0∑a​Car​xa−rΔxr−Ca0​xa​

= lim ⁡ Δ x → 0 ∑ r = 1 a C a r x a − r Δ x r Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\sum\limits_{r=1}^{a}{C_{a}^{r}x^{a-r}\Delta x^{r}}}{\Delta x}} =Δx→0lim​Δxr=1∑a​Car​xa−rΔxr​

= lim ⁡ Δ x → 0 ∑ r = 2 a C a r x a − r Δ x r − 1 + C a 1 x a − 1 =\lim\limits_{\Delta x\to 0}{\sum\limits_{r=2}^{a}{C_{a}^{r}x^{a-r}\Delta x^{r-1}+C_{a}^{1}x^{a-1}}} =Δx→0lim​r=2∑a​Car​xa−rΔxr−1+Ca1​xa−1

= C a 1 x a − 1 =C_a^{1}x^{a-1} =Ca1​xa−1

= a x a − 1 =a x^{a-1} =axa−1

貌似可以推广到 a ∈ R a\in \R a∈R 的情况，待完善

证明： f ′ ( x ) f'(x) f′(x)

= lim ⁡ Δ x → 0 sin ⁡ ( x + Δ x ) − sin ⁡ x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\sin (x+\Delta x)-\sin x}{\Delta x}} =Δx→0lim​Δxsin(x+Δx)−sinx​

= lim ⁡ Δ x → 0 sin ⁡ x cos ⁡ Δ x + sin ⁡ Δ x cos ⁡ x − sin ⁡ x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\sin x \cos\Delta x+\sin \Delta x \cos x-\sin x}{\Delta x}} =Δx→0lim​ΔxsinxcosΔx+sinΔxcosx−sinx​

= lim ⁡ Δ x → 0 sin ⁡ x ( cos ⁡ Δ x − 1 ) + sin ⁡ Δ x cos ⁡ x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\sin x(\cos\Delta x-1)+\sin\Delta x\cos x}{\Delta x}} =Δx→0lim​Δxsinx(cosΔx−1)+sinΔxcosx​

= lim ⁡ Δ x → 0 sin ⁡ x [ ( 1 − 2 sin ⁡ 2 Δ x 2 ) − 1 ] + cos ⁡ x ( 2 sin ⁡ Δ x 2 cos ⁡ Δ x 2 ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\sin x\left[\left(1-2\sin^2\dfrac{\Delta x}{2}\right)-1\right]+\cos x\left(2\sin\dfrac{\Delta x}{2}\cos \dfrac{\Delta x}{2}\right)}{\Delta x}} =Δx→0lim​Δxsinx[(1−2sin22Δx​)−1]+cosx(2sin2Δx​cos2Δx​)​

= lim ⁡ Δ x → 0 2 sin ⁡ Δ x 2 cos ⁡ Δ x 2 cos ⁡ x − 2 sin ⁡ 2 Δ x 2 sin ⁡ x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{2\sin\dfrac{\Delta x}{2}\cos\dfrac{\Delta x}{2}\cos x-2\sin^2\dfrac{\Delta x}{2}\sin x}{\Delta x}} =Δx→0lim​Δx2sin2Δx​cos2Δx​cosx−2sin22Δx​sinx​

= lim ⁡ Δ x → 0 2 sin ⁡ Δ x 2 ( cos ⁡ Δ x 2 cos ⁡ x − sin ⁡ x sin ⁡ Δ x 2 ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{2\sin\dfrac{\Delta x}{2}\left(\cos\dfrac{\Delta x}{2}\cos x-\sin x\sin \dfrac{\Delta x}{2}\right)}{\Delta x}} =Δx→0lim​Δx2sin2Δx​(cos2Δx​cosx−sinxsin2Δx​)​

= lim ⁡ Δ x → 0 2 sin ⁡ Δ x 2 cos ⁡ ( Δ x 2 + x ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{2\sin\dfrac{\Delta x}{2}\cos\left(\dfrac{\Delta x}{2}+x\right)}{\Delta x}} =Δx→0lim​Δx2sin2Δx​cos(2Δx​+x)​

= lim ⁡ Δ x → 0 cos ⁡ ( Δ x 2 + x ) sin ⁡ Δ x 2 Δ x 2 =\lim\limits_{\Delta x\to 0}{\cos\left(\dfrac{\Delta x}{2}+x\right)\dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}} =Δx→0lim​cos(2Δx​+x)2Δx​sin2Δx​​

= lim ⁡ Δ x → 0 cos ⁡ ( Δ x 2 + x ) =\lim\limits_{\Delta x\to 0}\cos\left(\dfrac{\Delta x}{2}+x\right) =Δx→0lim​cos(2Δx​+x)

= cos ⁡ x =\cos x =cosx

证明： f ′ ( x ) f'(x) f′(x)

= lim ⁡ Δ x → 0 cos ⁡ ( x + Δ x ) − cos ⁡ x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\cos(x+\Delta x)-\cos x}{\Delta x}} =Δx→0lim​Δxcos(x+Δx)−cosx​

= lim ⁡ Δ x → 0 cos ⁡ x cos ⁡ Δ x − sin ⁡ Δ x sin ⁡ x − cos ⁡ x Δ x =\lim\limits_{\Delta x\to 0}\dfrac{\cos x\cos \Delta x- \sin \Delta x\sin x-\cos x}{\Delta x} =Δx→0lim​ΔxcosxcosΔx−sinΔxsinx−cosx​

= lim ⁡ Δ x → 0 cos ⁡ x ( cos ⁡ Δ x − 1 ) − sin ⁡ Δ x sin ⁡ x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\cos x(\cos\Delta x-1)-\sin \Delta x\sin x}{\Delta x}} =Δx→0lim​Δxcosx(cosΔx−1)−sinΔxsinx​

= lim ⁡ Δ x → 0 cos ⁡ x [ ( 1 − 2 sin ⁡ 2 Δ x 2 ) − 1 ] − sin ⁡ x ( 2 sin ⁡ Δ x 2 cos ⁡ Δ x 2 ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\cos x \left[\left(1-2\sin^2\dfrac{\Delta x}{2}\right)-1\right]-\sin x\left(2\sin\dfrac{\Delta x}{2}\cos\dfrac{\Delta x}{2}\right)}{\Delta x}} =Δx→0lim​Δxcosx[(1−2sin22Δx​)−1]−sinx(2sin2Δx​cos2Δx​)​

= lim ⁡ Δ x → 0 − 2 sin ⁡ 2 Δ x 2 cos ⁡ x − 2 sin ⁡ Δ x 2 sin ⁡ x cos ⁡ Δ x 2 Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{-2\sin^2\dfrac{\Delta x}{2}\cos x-2\sin\dfrac{\Delta x}{2}\sin x \cos\dfrac{\Delta x}{2}}{\Delta x}} =Δx→0lim​Δx−2sin22Δx​cosx−2sin2Δx​sinxcos2Δx​​

= lim ⁡ Δ x → 0 − 2 sin ⁡ Δ x 2 ( sin ⁡ Δ x 2 cos ⁡ x + sin ⁡ x cos ⁡ Δ x 2 ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{-2\sin\dfrac{\Delta x}{2}\left(\sin \dfrac{\Delta x}{2}\cos x+\sin x \cos \dfrac{\Delta x}{2}\right)}{\Delta x}} =Δx→0lim​Δx−2sin2Δx​(sin2Δx​cosx+sinxcos2Δx​)​

= lim ⁡ Δ x → 0 − 2 sin ⁡ Δ x 2 sin ⁡ ( x + Δ x 2 ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{-2\sin \dfrac{\Delta x}{2}\sin\left(x+\dfrac{\Delta x}{2}\right)}{\Delta x}} =Δx→0lim​Δx−2sin2Δx​sin(x+2Δx​)​

= lim ⁡ Δ x → 0 − sin ⁡ ( x + Δ x 2 ) sin ⁡ Δ x 2 Δ x 2 =\lim\limits_{\Delta x\to 0}{-\sin\left(x+\dfrac{\Delta x}{2}\right)\dfrac{\sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}} =Δx→0lim​−sin(x+2Δx​)2Δx​sin2Δx​​

= lim ⁡ Δ x → 0 − sin ⁡ ( x + Δ x 2 ) =\lim\limits_{\Delta x\to 0}{-\sin\left(x+\dfrac{\Delta x}{2}\right)} =Δx→0lim​−sin(x+2Δx​)

= − sin ⁡ x =-\sin x =−sinx

证明： f ′ ( x ) f'(x) f′(x)

= lim ⁡ Δ x → 0 a x + Δ x − a x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{a^{x+\Delta x}-a^{x}}{\Delta x}} =Δx→0lim​Δxax+Δx−ax​

= lim ⁡ Δ x → 0 a x ( a Δ x − 1 ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{a^x(a^{\Delta x}-1)}{\Delta x}} =Δx→0lim​Δxax(aΔx−1)​

令 t = a Δ x − 1 t=a^{\Delta x}-1 t=aΔx−1 ，则 Δ x = log ⁡ a ( t + 1 ) \Delta x = \log_a(t+1) Δx=loga​(t+1)

∵ Δ x → 0 \because\Delta x\to 0 ∵Δx→0

∴ a Δ x → 1 \therefore a^{\Delta x}\to 1 ∴aΔx→1

∴ t + 1 → 1 \therefore t+1\to1 ∴t+1→1

即 t → 0 t\to0 t→0

∴ \therefore ∴ 原极限可化为

f ′ ( x ) f'(x) f′(x)

= lim ⁡ t → 0 a x t log ⁡ a ( t + 1 ) =\lim\limits_{t\to 0}{\dfrac{a^xt}{\log_a(t+1)}} =t→0lim​loga​(t+1)axt​

= lim ⁡ t → 0 a x 1 t log ⁡ a ( t + 1 ) =\lim\limits_{t\to 0}{\dfrac{a^x}{\dfrac{1}{t}\log_a(t+1)}} =t→0lim​t1​loga​(t+1)ax​

= lim ⁡ t → 0 a x log ⁡ a ( 1 + t ) 1 t =\lim\limits_{t\to 0}{\dfrac{a^x}{\log_a(1+t)^{\frac{1}{t}}}} =t→0lim​loga​(1+t)t1​ax​

= a x 1 log ⁡ a e =a^x\dfrac{1}{\log_ae} =axloga​e1​

= a x ln ⁡ a ln ⁡ e =a^x\dfrac{\ln a}{\ln e} =axlnelna​

= a x ln ⁡ a =a^x\ln a =axlna

特别地，当 a = e a=e a=e 时，即 f ( x ) = e x f(x)=e^x f(x)=ex ，有 f ′ ( x ) = e x f'(x)=e^x f′(x)=ex

证明： f ′ ( x ) f'(x) f′(x)

= lim ⁡ Δ x → 0 log ⁡ a ( x + Δ x ) − log ⁡ a x Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\log_a(x+\Delta x)-\log_ax}{\Delta x}} =Δx→0lim​Δxloga​(x+Δx)−loga​x​

= lim ⁡ Δ x → 0 log ⁡ a ( 1 + Δ x x ) Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\log_a\left(1+\dfrac{\Delta x}{x}\right)}{\Delta x}} =Δx→0lim​Δxloga​(1+xΔx​)​

= lim ⁡ Δ x → 0 log ⁡ a ( 1 + Δ x x ) 1 Δ x =\lim\limits_{\Delta x\to 0}{\log_a{\left(1+\dfrac{\Delta x}{x}\right)^{\frac{1}{\Delta x}}}} =Δx→0lim​loga​(1+xΔx​)Δx1​

令 t = Δ x x t=\dfrac{\Delta x}{x} t=xΔx​ ，则 1 Δ x = 1 t x \dfrac{1}{\Delta x} = \dfrac{1}{tx} Δx1​=tx1​

∵ Δ x → 0 \because \Delta x\to 0 ∵Δx→0

∴ Δ x x → 0 \therefore \dfrac{\Delta x}{x}\to 0 ∴xΔx​→0

∴ t → 0 \therefore t\to 0 ∴t→0

∴ \therefore ∴ 原极限可化为

f ′ ( x ) f'(x) f′(x)

= lim ⁡ t → 0 log ⁡ a ( 1 + t ) 1 t x =\lim\limits_{t\to 0}{\log_a\left(1+t\right)^{\frac{1}{tx}}} =t→0lim​loga​(1+t)tx1​

= lim ⁡ t → 0 1 x log ⁡ a ( 1 + t ) 1 t =\lim\limits_{t\to 0}{\dfrac{1}{x}\log_a\left(1+t\right)^{\frac{1}{t}}} =t→0lim​x1​loga​(1+t)t1​

= 1 x log ⁡ a e =\dfrac{1}{x}\log_ae =x1​loga​e

= 1 ln ⁡ e x ln ⁡ a =\dfrac{1\ln e}{x\ln a} =xlna1lne​

= 1 x ln ⁡ a =\dfrac{1}{x\ln a} =xlna1​

特别地，当 a = e a=e a=e 时，即 f ( x ) = ln ⁡ x f(x)=\ln x f(x)=lnx ，有 f ′ ( x ) = 1 x f'(x)=\dfrac{1}{x} f′(x)=x1​

证明： f ′ ( x ) f'(x) f′(x)

= lim ⁡ Δ x → 0 ϕ − ϕ Δ x =\lim\limits_{\Delta x\to 0}{\dfrac{\phi-\phi}{\Delta x}} =Δx→0lim​Δxϕ−ϕ​

= 0 =0 =0

本文推导了部分基本的导数公式

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