4. 微分
4.3 导数四则运算与反函数求导法则
通过例题，计算常用的基本初等函数的导数
【例4.3.1】 y = sin x y=\sin x y=sinx
【解】 y ′ ( x ) = lim Δ x → 0 sin ( x + Δ x ) − sin x Δ x = lim Δ x → 0 2 cos 2 x + Δ x 2 sin Δ x 2 Δ x = cos x y'(x)=\lim\limits_{\Delta x\to 0}\frac{\sin (x+\Delta x)-\sin x}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{2\cos\frac{2x+\Delta x}{2}\sin\frac{\Delta x}{2}}{\Delta x}= \cos x y′(x)=Δx→0limΔxsin(x+Δx)−sinx=Δx→0limΔx2cos22x+Δxsin2Δx=cosx
同理 y = cos x , y ′ = − sin x y=\cos x,y' = -\sin x y=cosx,y′=−sinx
【例4.3.2】 y = ln x y=\ln x y=lnx
【解】 y ′ ( x ) = lim Δ x → 0 ln ( x + Δ x ) − ln x Δ x = lim Δ x → 0 ln ( 1 + Δ x x ) Δ x = lim Δ x → 0 Δ x x Δ x = 1 x y'(x)=\lim\limits_{\Delta x\to 0}\frac{\ln(x+\Delta x)-\ln x}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{\ln(1+\frac{\Delta x}{x})}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{\frac{\Delta x}{x}}{\Delta x}=\frac{1}{x} y′(x)=Δx→0limΔxln(x+Δx)−lnx=Δx→0limΔxln(1+xΔx)=Δx→0limΔxxΔx=x1
【例4.3.3】 y = e x y=e^x y=ex
【解】 y ′ ( x ) = lim Δ x → 0 e x + Δ x − e x Δ x = lim Δ x → 0 e x ( e Δ x − 1 ) Δ x = lim Δ x → 0 e x Δ x Δ x = e x y'(x)=\lim\limits_{\Delta x\to 0}\frac{e^{x+\Delta x}-e^x}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^x(e^{\Delta x}-1)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^x\Delta x}{\Delta x}=e^x y′(x)=Δx→0limΔxex+Δx−ex=Δx→0limΔxex(eΔx−1)=Δx→0limΔxexΔx=ex
【注】 y = a x = e ln a x = e x ln a , y ′ ( x ) = lim Δ x → 0 e ( x + Δ x ) ln a − e x ln a Δ x = lim Δ x → 0 e x ln a ( e Δ x ln a − 1 ) Δ x = lim Δ x → 0 e x ln a Δ x ln a Δ x = e x ln a ln a = a x ln a y=a^x=e^{\ln a^x}=e^{x\ln a},y'(x)=\lim\limits_{\Delta x\to 0}\frac{e^{(x+\Delta x)\ln a}-e^{x\ln a}}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^{x\ln a}(e^{\Delta x\ln a}-1)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{e^{x\ln a}\Delta x\ln a}{\Delta x}=e^{x\ln a}\ln a=a^x\ln a y=ax=elnax=exlna,y′(x)=Δx→0limΔxe(x+Δx)lna−exlna=Δx→0limΔxexlna(eΔxlna−1)=Δx→0limΔxexlnaΔxlna=exlnalna=axlna
【例4.3.4】 y = x α , α ∈ R , x > 0 y=x^{\alpha},\alpha\in\mathbb{R},x>0 y=xα,α∈R,x>0
【解】 y ′ ( x ) = lim Δ x → 0 ( x + Δ x ) α − x α Δ x = lim Δ x → 0 x α ( ( 1 + Δ x x ) α − 1 ) Δ x = lim Δ x → 0 x α ⋅ α Δ x x Δ x = α x α − 1 y'(x)=\lim\limits_{\Delta x\to 0}\frac{(x+\Delta x)^{\alpha}-x^\alpha}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{x^\alpha ((1+\frac{\Delta x}{x})^\alpha - 1)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{x^\alpha \cdot \alpha\frac{\Delta x}{x}}{\Delta x}=\alpha x^{\alpha - 1} y′(x)=Δx→0limΔx(x+Δx)α−xα=Δx→0limΔxxα((1+xΔx)α−1)=Δx→0limΔxxα⋅αxΔx=αxα−1
【注】 y = x n , n = 1 , 2 , 3 , . . . , x ∈ ( − ∞ , + ∞ ) , y ′ ( x ) = n x n − 1 , x ∈ ( − ∞ , + ∞ ) y=x^n,n=1,2,3,...,x\in (-\infty,+\infty),y'(x)=nx^{n-1},x\in(-\infty,+\infty) y=xn,n=1,2,3,...,x∈(−∞,+∞),y′(x)=nxn−1,x∈(−∞,+∞)
y = x − n , n = 1 , 2 , 3 , . . . , x ∈ ( − ∞ , 0 ) ∪ ( 0 , + ∞ ) , y ′ ( x ) = − n x − n − 1 = − n x n + 1 , x ∈ ( − ∞ , 0 ) ∪ ( 0 , + ∞ ) y=x^{-n},n=1,2,3,...,x\in(-\infty,0)\cup(0,+\infty),y'(x)=-nx^{-n-1}=\frac{-n}{x^{n+1}},x\in(-\infty,0)\cup(0,+\infty) y=x−n,n=1,2,3,...,x∈(−∞,0)∪(0,+∞),y′(x)=−nx−n−1=xn+1−n,x∈(−∞,0)∪(0,+∞)
【例】 y = x 2 3 , x ∈ ( − ∞ , + ∞ ) , y ′ ( x ) = 2 3 x − 1 3 = 2 3 x 3 , x ∈ ( − ∞ , 0 ) ∪ ( 0 , + ∞ ) y=x^{\frac{2}{3}},x\in(-\infty,+\infty),y'(x)=\frac{2}{3}x^{-\frac{1}{3}}=\frac{2}{3\sqrt[3]{x}},x\in(-\infty,0)\cup(0,+\infty) y=x32,x∈(−∞,+∞),y′(x)=32x−31=33x 2,x∈(−∞,0)∪(0,+∞)，即它在 0 0 0点不可导。
【例】 y = x 1 2 , x ∈ [ 0 , + ∞ ) , y ′ ( x ) = 1 2 x , x ∈ ( 0 , + ∞ ) y=x^{\frac{1}{2}},x\in[0,+\infty),y'(x)=\frac{1}{2\sqrt{x}},x\in(0,+\infty) y=x21,x∈[0,+∞),y′(x)=2x 1,x∈(0,+∞)，它在 0 0 0点右导数不存在。
4.3.1 导数四则运算
【定理4.3.1】 f f f和 g g g在同一区间可导，则 c 1 f ( x ) + c 2 g ( x ) c_1f(x)+c_2g(x) c1f(x)+c2g(x)也在该区间可导。且有 ( c 1 f ( x ) + c 2 g ( x ) ) ′ = c 1 f ′ ( x ) + c 2 g ′ ( x ) (c_1f(x)+c_2g(x))'=c_1f'(x)+c_2g'(x) (c1f(x)+c2g(x))′=c1f′(x)+c2g′(x)
【证】 ( c 1 f ( x ) + c 2 g ( x ) ) ′ = lim Δ x → 0 ( c 1 f ( x + Δ x ) + c 2 g ( x + Δ x ) ) − ( c 1 f ( x ) + c 2 g ( x ) ) Δ x = lim Δ x → 0 ( c 1 f ( x + Δ x ) − f ( x ) Δ x + c 2 g ( x + Δ x ) − g ( x ) Δ x ) = c 1 f ′ ( x ) + c 2 g ′ ( x ) (c_1f(x)+c_2g(x))'=\lim\limits_{\Delta x\to 0}\frac{(c_1f(x+\Delta x)+c_2g(x+\Delta x))-(c_1f(x)+c_2g(x))}{\Delta x}=\lim\limits_{\Delta x\to 0}(c_1\frac{f(x+\Delta x)-f(x)}{\Delta x}+c_2\frac{g(x+\Delta x)-g(x)}{\Delta x})=c_1f'(x)+c_2g'(x) (c1f(x)+c2g(x))′=Δx→0limΔx(c1f(x+Δx)+c2g(x+Δx))−(c1f(x)+c2g(x))=Δx→0lim(c1Δxf(x+Δx)−f(x)+c2Δxg(x+Δx)−g(x))=c1f′(x)+c2g′(x)
【例】 ( log a x ) ′ = ( ln x ln a ) ′ = 1 ln a ⋅ 1 x = 1 x ln a (\log_{a}x)'=(\frac{\ln x}{\ln a})'=\frac{1}{\ln a}\cdot\frac{1}{x}=\frac{1}{x\ln a} (logax)′=(lnalnx)′=lna1⋅x1=xlna1
【定理4.3.2】 f f f和 g g g在同一区间可导，则 f ( x ) ⋅ g ( x ) f(x)\cdot g(x) f(x)⋅g(x)也在该区间可导， ( f ( x ) 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))′=f′(x)g(x)+f(x)g′(x)
【证】 ( f ( x ) g ( x ) ) ′ = lim Δ x → 0 f ( x + Δ x ) g ( x + Δ x ) − f ( x ) g ( x ) Δ x = lim Δ x → 0 f ( x + Δ x ) g ( x + Δ x ) − f ( x + Δ x ) g ( x ) + f ( x + Δ x ) g ( x ) − f ( x ) g ( x ) Δ x = lim Δ x → 0 f ( x + Δ x ) ( g ( x + Δ x ) − g ( x ) ) Δ x + lim Δ x → 0 g ( x ) ( f ( x + Δ x ) − f ( x ) ) Δ x = ( f 可导必连续 ) f ( x ) g ′ ( x ) + g ( x ) f ′ ( x ) = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) (f(x)g(x))'=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x)g(x)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)g(x+\Delta x)-f(x+\Delta x)g(x)+f(x+\Delta x)g(x)-f(x)g(x)}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{f(x+\Delta x)(g(x+\Delta x)-g(x))}{\Delta x}+\lim\limits_{\Delta x\to 0}\frac{g(x)(f(x+\Delta x)-f(x))}{\Delta x}=(f可导必连续)f(x)g'(x)+g(x)f'(x)=f'(x)g(x)+f(x)g'(x) (f(x)g(x))′=Δx→0limΔxf(x+Δx)g(x+Δx)−f(x)g(x)=Δx→0limΔxf(x+Δx)g(x+Δx)−f(x+Δx)g(x)+f(x+Δx)g(x)−f(x)g(x)=Δx→0limΔxf(x+Δx)(g(x+Δx)−g(x))+Δx→0limΔxg(x)(f(x+Δx)−f(x))=(f可导必连续)f(x)g′(x)+g(x)f′(x)=f′(x)g(x)+f(x)g′(x)
【例4.3.6】求 y = 3 x cos x y=3^{x}\cos x y=3xcosx的导数
【解】 y ′ ( x ) = 3 x ( ln 3 ) ⋅ cos x − 3 x sin x = 3 x ( ( ln 3 ) ⋅ cos x − sin x ) y'(x)=3^x(\ln 3)\cdot\cos x-3^x\sin x=3^x((\ln 3)\cdot \cos x - \sin x) y′(x)=3x(ln3)⋅cosx−3xsinx=3x((ln3)⋅cosx−sinx)
【例4.3.7】求 y = sin x x y=\frac{\sin x}{x} y=xsinx的导数。
【解】 y ′ ( x ) = 1 x cos x + ( − 1 x 2 ) sin x = x cos x − sin x x 2 y'(x)=\frac{1}{x}\cos x+(-\frac{1}{x^2})\sin x=\frac{x\cos x - \sin x}{x^2} y′(x)=x1cosx+(−x21)sinx=x2xcosx−sinx
【定理4.3.3】设 g ( x ) g(x) g(x)在某一个区间可导， g ( x ) ≠ 0 g(x)\ne 0 g(x)=0，则 1 g ( x ) \frac{1}{g(x)} g(x)1也在该区间可导，并且 ( 1 g ( x ) ) ′ = − g ′ ( x ) g 2 ( x ) (\frac{1}{g(x)})'=\frac{-g'(x)}{g^{2}(x)} (g(x)1)′=g2(x)−g′(x)
【证】 ( 1 g ( x ) ) ′ = lim Δ x → 0 1 g ( x + Δ x ) − 1 g ( x ) Δ x = lim Δ x → 0 − ( g ( x + Δ x ) − g ( x ) ) Δ x g ( x ) g ( x + Δ x ) = ( g 可导必连续 ) − g ′ ( x ) g 2 ( x ) (\frac{1}{g(x)})'=\lim\limits_{\Delta x\to 0}\frac{\frac{1}{g(x+\Delta x)}-\frac{1}{g(x)}}{\Delta x}=\lim\limits_{\Delta x\to 0}\frac{-(g(x+\Delta x)-g(x))}{\Delta xg(x)g(x+\Delta x)}=(g可导必连续)\frac{-g'(x)}{g^{2}(x)} (g(x)1)′=Δx→0limΔxg(x+Δx)1−g(x)1=Δx→0limΔxg(x)g(x+Δx)−(g(x+Δx)−g(x))=(g可导必连续)g2(x)−g′(x)
【注】 d f = f ′ ( x ) d x ⇔ d y d x = f ′ ( x ) , d ( 1 g ( x ) ) = − g ′ ( x ) g 2 ( x ) d x = − d g ( x ) g 2 ( x ) df=f'(x)dx\Leftrightarrow \frac{dy}{dx}=f'(x),d(\frac{1}{g(x)})=\frac{-g'(x)}{g^{2}(x)}dx=-\frac{dg(x)}{g^{2}(x)} df=f′(x)dx⇔dxdy=f′(x),d(g(x)1)=g2(x)−g′(x)dx=−g2(x)dg(x)
【例】 ( sin x ) ′ = cos x , ( cos x ) ′ = − sin x (\sin x)'=\cos x,(\cos x)'=-\sin x (sinx)′=cosx,(cosx)′=−sinx，求 ( sec x ) ′ (\sec x)' (secx)′
【解】 ( sec x ) ′ = ( 1 cos x ) ′ = − ( − sin x ) cos 2 x = sin x cos 2 x = tan x sec x (\sec x)'=(\frac{1}{\cos x})'=-\frac{(-\sin x)}{\cos^2 x}=\frac{\sin x}{\cos ^ 2 x}=\tan x\sec x (secx)′=(cosx1)′=−cos2x(−sinx)=cos2xsinx=tanxsecx
同理 ( csc x ) ′ = − cot x csc x (\csc x)'=-\cot x\csc x (cscx)′=−cotxcscx
【推论】 f , g f,g f,g在同一区间可导， g ( x ) ≠ 0 g(x)\ne 0 g(x)=0，则 f ( x ) g ( x ) \frac{f(x)}{g(x)} g(x)f(x)在该区间可导，并且 ( f ( x ) g ( x ) ) ′ = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g 2 ( x ) (\frac{f(x)}{g(x)})'=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} (g(x)f(x))′=g2(x)f′(x)g(x)−f(x)g′(x)
【证】可以由前面的两个公式推出
( f ( x ) g ( x ) ) ′ = ( f ( x ) ⋅ 1 g ( x ) ) ′ = f ′ ( x ) 1 g ( x ) + f ( x ) ⋅ ( − g ′ ( x ) g 2 ( x ) ) = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g 2 ( x ) (\frac{f(x)}{g(x)})'=(f(x)\cdot\frac{1}{g(x)})'=f'(x)\frac{1}{g(x)}+f(x)\cdot(-\frac{g'(x)}{g^2(x)})=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)} (g(x)f(x))′=(f(x)⋅g(x)1)′=f′(x)g(x)1+f(x)⋅(−g2(x)g′(x))=g2(x)f′(x)g(x)−f(x)g′(x)
【例】 ( tan x ) ′ = ( sin x cos x ) ′ = cos 2 x − ( sin x ⋅ ( − sin x ) ) cos 2 x = cos 2 x + sin 2 x cos 2 x = 1 cos 2 x = sec 2 x (\tan x)'=(\frac{\sin x}{\cos x})'=\frac{\cos^2 x-(\sin x\cdot (-\sin x))}{\cos ^2 x}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2 x (tanx)′=(cosxsinx)′=cos2xcos2x−(sinx⋅(−sinx))=cos2xcos2x+sin2x=cos2x1=sec2x
同理 ( cot x ) ′ = − csc 2 x (\cot x)'=-\csc^ 2 x (cotx)′=−csc2x
4.3.2 反函数求导法则
【定理4.3.4】【反函数求导定理】 f ( x ) f(x) f(x)在 ( a , b ) (a,b) (a,b)连续且严格单调并且可导， f ′ ( x ) ≠ 0 f'(x)\ne 0 f′(x)=0， α = min { f ( a + ) , f ( b − ) } , β = { f ( a + ) , f ( b − ) } \alpha =\min\{f(a+),f(b-)\},\beta=\{f(a+),f(b-)\} α=min{f(a+),f(b−)},β={f(a+),f(b−)}，则 f − 1 ( y ) f^{-1}(y) f−1(y)在 ( α , β ) (\alpha,\beta) (α,β)上可导且 ( f − 1 ( y ) ) ′ = 1 f ′ ( x ) , ( x = f − 1 ( y ) ) (f^{-1}(y))'=\frac{1}{f'(x)},(x=f^{-1}(y)) (f−1(y))′=f′(x)1,(x=f−1(y))
【证】 y = f ( x ) , y + Δ y = f ( x + Δ x ) , x = f − 1 ( y ) , f − 1 ( y + Δ y ) = x + Δ x y=f(x),y+\Delta y=f(x+\Delta x),x=f^{-1}(y),f^{-1}(y+\Delta y)=x+\Delta x y=f(x),y+Δy=f(x+Δx),x=f−1(y),f−1(y+Δy)=x+Δx
由于 f f f是严格单调的， Δ x ≠ 0 ⇔ Δ y ≠ 0 \Delta x\ne 0\Leftrightarrow \Delta y \ne 0 Δx=0⇔Δy=0，由 f f f的连续性， Δ x → 0 ⇔ Δ y → 0 \Delta x\to 0\Leftrightarrow\Delta y\to 0 Δx→0⇔Δy→0
实际上 ( f − 1 ( y ) ) ′ = lim Δ y → 0 f − 1 ( y + Δ y ) − f − 1 ( y ) Δ y = lim Δ y → 0 x + Δ x − x Δ y = lim Δ y → 0 Δ x f ( x + Δ x ) − y = lim Δ x → 0 Δ x f ( x + Δ x ) − f ( x ) = lim Δ x → 0 1 f ( x + Δ x ) − f ( x ) Δ x = 1 f ′ ( x ) (f^{-1}(y))'=\lim\limits_{\Delta y\to 0}\frac{f^{-1}(y+\Delta y)-f^{-1}(y)}{\Delta y}=\lim\limits_{\Delta y\to 0}\frac{x + \Delta x - x}{\Delta y}=\lim\limits_{\Delta y\to 0}\frac{\Delta x}{f(x+\Delta x)-y}=\lim\limits_{\Delta x\to 0}\frac{\Delta x}{f(x+\Delta x)-f(x)}=\lim\limits_{\Delta x\to 0}\frac{1}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}=\frac{1}{f'(x)} (f−1(y))′=Δy→0limΔyf−1(y+Δy)−f−1(y)=Δy→0limΔyx+Δx−x=Δy→0limf(x+Δx)−yΔx=Δx→0limf(x+Δx)−f(x)Δx=Δx→0limΔxf(x+Δx)−f(x)1=f′(x)1（因为 Δ x \Delta x Δx趋于0但是不等于0，可以除下来）
【例】 ( arctan x ) ′ = 1 ( tan y ) ′ = 1 sec 2 y = 1 1 + tan 2 y = 1 1 + x 2 , ( y = arctan x , tan y = x ) (\arctan x)'=\frac{1}{(\tan y)'}=\frac{1}{\sec^2 y}=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2},(y=\arctan x,\tan y =x) (arctanx)′=(tany)′1=sec2y1=1+tan2y1=1+x21,(y=arctanx,tany=x)
同理 ( arccot x ) ′ = − 1 1 + x 2 (\text{arccot} x)'=-\frac{1}{1+x^2} (arccotx)′=−1+x21
【例】 ( arcsin x ) ′ = 1 ( sin y ) ′ = 1 cos y = 1 1 − sin 2 y = 1 1 − x 2 , ( x = sin y ) (\arcsin x)'=\frac{1}{(\sin y)'}=\frac{1}{\cos y}=\frac{1}{\sqrt{1-\sin^2 y}}=\frac{1}{\sqrt{1-x^2}},(x=\sin y) (arcsinx)′=(siny)′1=cosy1=1−sin2y 1=1−x2 1,(x=siny)
同理 ( arccos x ) ′ = − 1 1 − x 2 (\arccos x)'=-\frac{1}{\sqrt{1-x^2}} (arccosx)′=−1−x2 1