考研数学：常见的初等函数求导公式以及其对应的积分公式

( x u ) ′ = μ x k − 1 ∫ μ x n − 1 d x = x μ + c \left(x^{u}\right)^{\prime}=\mu x^{k-1} \quad \quad \int \mu x^{n-1} \mathrm{d} x=x^{\mu}+c (xu)′=μxk−1∫μxn−1dx=xμ+c

( x m p ) ′ = m − p p x m p ∫ m p x m − p p d x = x m p = x m p + c (\sqrt[p]{x^{m}})^{\prime}=\frac{m-p}{p} x^{\frac{m}{p}} \quad \int \frac{m}{p} x^{\frac{m-p}{p}} \mathrm{d} x=x^{\frac{m}{p}}=\sqrt[p]{x^{m}}+c (pxm ​)′=pm−p​xpm​∫pm​xpm−p​dx=xpm​=pxm ​+c

( ln ⁡ ∣ x ∣ ) ′ = 1 x ∫ 1 x d x = ln ⁡ ∣ x ∣ + c (\ln |x|)^{\prime}=\frac{1}{x} \quad \int \frac{1}{x} \mathrm{d} x=\ln |x|+c (ln∣x∣)′=x1​∫x1​dx=ln∣x∣+c

( e x ) ′ = e x ∫ e x d x = e x + c \left(\mathrm{e}^{x}\right)^{\prime}=\mathrm{e}^{x} \quad \int \mathrm{e}^{x} \mathrm{d} x=\mathrm{e}^{x}+c (ex)′=ex∫exdx=ex+c

( a x ) ′ = a x ln ⁡ a ∫ a x d x = a x ln ⁡ a + c \left(a^{x}\right)^{\prime}=a^{x} \ln a \quad \int a^{x} \mathrm{d} x=\frac{a^{x}}{\ln a}+c (ax)′=axlna∫axdx=lnaax​+c

( sin ⁡ x ) ′ = cos ⁡ x ∫ cos ⁡ x d x = sin ⁡ x + c (\sin x)^{\prime}=\cos x \quad \int \cos x d x=\sin x+c (sinx)′=cosx∫cosxdx=sinx+c

( cos ⁡ x ) ′ = − sin ⁡ x ∫ sin ⁡ x d x = − cos ⁡ x + c (\cos x)^{\prime}=-\sin x \quad \int \sin x d x=-\cos x+c (cosx)′=−sinx∫sinxdx=−cosx+c

( tan ⁡ x ) ′ = sec ⁡ 2 x ∫ sec ⁡ 2 x d x = tan ⁡ x + c (\tan x)^{\prime}=\sec ^{2} x \quad \int \sec ^{2} x d x=\tan x+c (tanx)′=sec2x∫sec2xdx=tanx+c

( cot ⁡ x ) ′ = − csc ⁡ 2 x ∫ csc ⁡ 2 x d x = − cot ⁡ x + c (\cot x)^{\prime}=-\csc ^{2} x \quad \int \csc ^{2} x d x=-\cot x+c (cotx)′=−csc2x∫csc2xdx=−cotx+c

( sec ⁡ x ) ′ = sec ⁡ x tan ⁡ x ∫ sec ⁡ x tan ⁡ x d x = sec ⁡ x + c (\sec x)^{\prime}=\sec x \tan x \quad \int \sec x \tan x d x=\sec x+c (secx)′=secxtanx∫secxtanxdx=secx+c

( csc ⁡ x ) ′ = − csc ⁡ x cot ⁡ x ∫ csc ⁡ x cot ⁡ x d x = − csc ⁡ x + c (\csc x)^{\prime}=-\csc x \cot x \quad \int \csc x \cot x d x=-\csc x+c (cscx)′=−cscxcotx∫cscxcotxdx=−cscx+c

( arcsin ⁡ x ) ′ = 1 1 − x 2 ∫ 1 1 − x 2 d x = arcsin ⁡ x + c (\arcsin x)^{\prime}=\frac{1}{\sqrt{1-x^{2}}} \quad \int \frac{1}{\sqrt{1-x^{2}}} \mathrm{d} x=\arcsin x+c (arcsinx)′=1−x2 ​1​∫1−x2 ​1​dx=arcsinx+c

( arccos ⁡ x ) ′ = − 1 1 − x 2 ∫ 1 1 − x 2 d x = − arccos ⁡ x + c (\arccos x)^{\prime}=-\frac{1}{\sqrt{1-x^{2}}} \quad \int \frac{1}{\sqrt{1-x^{2}}} \mathrm{d} x=-\arccos x+c (arccosx)′=−1−x2 ​1​∫1−x2 ​1​dx=−arccosx+c

( arctan ⁡ x ) ′ = 1 1 + x 2 ∫ 1 1 + x 2 d x = arctan ⁡ x + c (\arctan x)^{\prime}=\frac{1}{1+x^{2}} \quad \int \frac{1}{1+x^{2}} \mathrm{d} x=\arctan x+c (arctanx)′=1+x21​∫1+x21​dx=arctanx+c

( arccot ⁡ x ) ′ = − 1 1 + x 2 ∫ 1 1 + x 2 d x = − arccot ⁡ x + c (\operatorname{arccot} x)^{\prime}=-\frac{1}{1+x^{2}} \quad \int \frac{1}{1+x^{2}} \mathrm{d} x=-\operatorname{arccot} x+c (arccotx)′=−1+x21​∫1+x21​dx=−arccotx+c

( arcsec ⁡ x ) ′ = 1 x x 2 − 1 ∫ 1 x x 2 − 1 d x = arcsec ⁡ x + c (\operatorname{arcsec} x)^{\prime}=\frac{1}{x \sqrt{x^{2}-1}} \quad \int \frac{1}{x \sqrt{x^{2}-1}} \mathrm{d} x=\operatorname{arcsec} x+c (arcsecx)′=xx2−1 ​1​∫xx2−1 ​1​dx=arcsecx+c

( arccsc ⁡ x ) ′ = − 1 x x 2 − 1 ∫ 1 x x 2 − 1 d x = − arccsc ⁡ x + c (\operatorname{arccsc} x)^{\prime}=-\frac{1}{x \sqrt{x^{2}-1}} \quad \int \frac{1}{x \sqrt{x^{2}-1}} \mathrm{d} x=-\operatorname{arccsc} x+c (arccscx)′=−xx2−1 ​1​∫xx2−1 ​1​dx=−arccscx+c

( ln ⁡ ∣ x + x 2 ± b ∣ ) ′ = 1 x 2 ± b ∫ 1 x 2 ± b d x = ln ⁡ ∣ x + x 2 ± b ∣ + c (\ln |x+\sqrt{x^{2} \pm b}|)^{\prime}=\frac{1}{\sqrt{x^{2} \pm b}} \quad \int \frac{1}{\sqrt{x^{2} \pm b}} \mathrm{d} x=\ln |x+\sqrt{x^{2} \pm b}|+c (ln∣x+x2±b ​∣)′=x2±b ​1​∫x2±b ​1​dx=ln∣x+x2±b ​∣+c

∫ sin ⁡ 2 x d x = x 2 − sin ⁡ 2 x 4 + C ( sin ⁡ 2 x = 1 − cos ⁡ 2 x 2 ) \int \sin ^{2} x \mathrm{d} x=\frac{x}{2}-\frac{\sin 2 x}{4}+C\left(\sin ^{2} x=\frac{1-\cos 2 x}{2}\right) ∫sin2xdx=2x​−4sin2x​+C(sin2x=21−cos2x​)

∫ cos ⁡ 2 x d x = x 2 + sin ⁡ 2 x 4 + C ( cos ⁡ 2 x = 1 + cos ⁡ 2 x 2 ) \int \cos ^{2} x \mathrm{d} x=\frac{x}{2}+\frac{\sin 2 x}{4}+C\left(\cos ^{2} x=\frac{1+\cos 2 x}{2}\right) ∫cos2xdx=2x​+4sin2x​+C(cos2x=21+cos2x​)

∫ tan ⁡ 2 x d x = tan ⁡ x − x + C ( tan ⁡ 2 x = sec ⁡ 2 x − 1 ) \int \tan ^{2} x \mathrm{d} x=\tan x-x+C\left(\tan ^{2} x=\sec ^{2} x-1\right) ∫tan2xdx=tanx−x+C(tan2x=sec2x−1)

∫ cot ⁡ 2 x d x = − cot ⁡ x − x + C ( cot ⁡ 2 x = csc ⁡ 2 x − 1 ) \int \cot ^{2} x \mathrm{d} x=-\cot x-x+C\left(\cot ^{2} x=\csc ^{2} x-1\right) ∫cot2xdx=−cotx−x+C(cot2x=csc2x−1)