導(dǎo)數(shù)16基本公式推1

1、導(dǎo)數(shù)16基本公式推導(dǎo)t(cy = th g為常數(shù))2 .(x+” 二 口- -。且a1)a xl na14 . (Inx)=- x5.13) = alna, (a 6且3 * 1)6. (J) = /7. (s in x) = cos x8. fcosx/ 二- s in x,210, (Cot X) = - C3C X11, (sec x) = sec xtan x12P (esc x)+ = - sec xcotx1 J. (arcs i n xj - |1 - x14. (arccos x)=15. arctan xY 二1 + X116. (arccot x)r =-1 * x9. (

2、tan x)h = sei xt (c),= 0, (c為常數(shù))設(shè)函數(shù)=f(x)= 為常數(shù)上顯然1 f5+ Ax)=c,則有Ay = f(x + &町-f(M) = c - c = 0HA y 0-7- - 0. o x x v I im 二 I in 0 = 0 10 k & 10即 = 02.（xa） 二 QX口 J （a 為實數(shù)）設(shè)函數(shù)y = /口為自然數(shù)），則有AxA yI in I i m iAx-0 * 汽X-*。n nxn(n - 1) n - 22Xin * 1dx + + ()()n nxri nx3. (Iogax5,=赤,(a)。且 a*1)A x設(shè)函數(shù)y = loax(

3、a 俎0* 1),則有Ay = log3(x + Ax) - logax - logil +14t (lnx)s =一x當口二焉中一甌則有a”y /5. (a*) = aK|n ar (a )。且a 豐 1)已知v = a*是X = Iq4V的反函數(shù)，而國=10”在(0, + 8)上單調(diào)可導(dǎo).且 fl1logdy)=豐 0.“ yin a故在對應(yīng)的區(qū)間(-81 + 8)內(nèi)有即k (ex), = J當(/) = a*In 1a中自二 e時，有(J) - e。7. (s i n x)h = cos x設(shè)函數(shù) = sin尤則有/ A A xAy = s i n (x + Ax) - s i n x

4、= 2co& |x + -jsA x s i n以t42=8邛 + 0A， x2 ) A xT y iI im I im (x +/3Iim cos x + -I2A xin -I im = cc$C3 X.8, (cos x)si nxA KAy- cos yA KA x-t- 8 in-2 )2A x sirt -Iim sinfx + 4m-02I im & -* D *s i nx.29. (tan x) = sec x(tan x)Pcos xi3inx)-cos x - s i nxicosx)-2 COS X2 COS X2 COS X2 sec xtan x.即tsecx)=

5、sec xtanx10. (cot x),二2CSC X(cot JC ).2 si n(cos x )Fs i n x - x(si n x)sin xGOS Xesc xsm xs in x即eotesc x11. (sec x),sec xtan xI GOS X )i COS X2COS X2COS X-secxtanx即tsecx)，= sec xtanx12, (esc x)k-CSC xcot X(CSC X,CSC KCOt Xsin k13, (arcs inx)=1 - x1(GOS v)=arecas x)F =s i n ja rccos x)1， =(arc in x)F =is in y)11-.in2(arc.inx) =即i且res in X)1=14, (arccosx)=11 - x2即 arccos x15* (arctan x)h1 + x2V - (arctan x/ =-=(tany) sec y sec (arctanx) 1 + tan (arctanx)yP11 + x即(Cretan =16. (arc