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)=Δx0limΔxsin(x+Δx)sinx=Δx0limΔx2cos22x+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)=Δx0limΔxln(x+Δx)lnx=Δx0limΔxln(1+xΔx)=Δx0limΔxxΔx=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)=Δx0limΔxex+Δxex=Δx0limΔxex(eΔx1)=Δx0limΔxexΔx=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)=Δx0limΔxe(x+Δx)lnaexlna=Δx0limΔxexlna(eΔxlna1)=Δx0limΔxexlnaΔxlna=exlnalna=axlna
【例4.3.4】 y = x α , α ∈ R , x > 0 y=x^{\alpha},\alpha\in\mathbb{R},x>0 y=xα,αR,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)=Δx0limΔx(x+Δx)αxα=Δx0limΔxxα((1+xΔx)α1)=Δx0limΔxxααxΔx=αxα1
【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)=nxn1,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=xn,n=1,2,3,...,x(,0)(0,+),y(x)=nxn1=xn+1n,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)=32x31=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 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))=Δx0limΔx(c1f(x+Δx)+c2g(x+Δx))(c1f(x)+c2g(x))=Δx0lim(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)=lna1x1=xlna1


【定理4.3.2】 f f f 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))=Δx0limΔxf(x+Δx)g(x+Δx)f(x)g(x)=Δx0limΔxf(x+Δx)g(x+Δx)f(x+Δx)g(x)+f(x+Δx)g(x)f(x)g(x)=Δx0limΔxf(x+Δx)(g(x+Δx)g(x))+Δx0limΔxg(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)cosx3xsinx=3x((ln3)cosxsinx)


【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=x2xcosxsinx


【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)=Δx0limΔxg(x+Δx)1g(x)1=Δx0limΔxg(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)dxdxdy=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) ( 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) f1(y) ( 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)) (f1(y))=f(x)1,(x=f1(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=f1(y),f1(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=0Δy=0,由 f f f的连续性, Δ x → 0 ⇔ Δ y → 0 \Delta x\to 0\Leftrightarrow\Delta y\to 0 Δx0Δy0
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)} (f1(y))=Δy0limΔyf1(y+Δy)f1(y)=Δy0limΔyx+Δxx=Δy0limf(x+Δx)yΔx=Δx0limf(x+Δx)f(x)Δx=Δx0limΔxf(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=1sin2y 1=1x2 1,(x=siny)
同理 ( arccos ⁡ x ) ′ = − 1 1 − x 2 (\arccos x)'=-\frac{1}{\sqrt{1-x^2}} (arccosx)=1x2 1

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