O HM V VI PHN.

Bi ton: Vn ng vin chy v bi phi hp.

O HM TI 1 IMCho y = f(x) xc nh trong (a, b) x0, xt t s Nu t s trn c gii hn hu hn khi x x0 hay x 0 th f c o hm ti x0.t

x x0f(x0) l h s gc tip tuyn ca ng cong (C): y = f(x) ti tip im M(x0, f(x0))x

o hm tri ti x0:o hm phi ti x0:f c o hm ti x0

Cch tnh o hmNu f xc nh bi 1 biu thc s cp: dng cng thc o hm v cc quy tc(tng, hiu, tch, thng, hm hp).Nu ti x0, biu thc f khng xc nh: tnh bng nh ngha.Nu hm s c phn chia biu thc ti x0: tnh bng nh ngha.Nu f(x) = u(x)v(x) hoc f(x) l tch thng ca nhiu hm: tnh (lnf)

V d: tnh o hm ti cc im c ch rati x = 1

ti x = 01x 0-1f (0) khng tn tix 0+

ti x = 1

o hm v lin tcVD: tm cc hng s a, b f c o hm ti x0(Nn xt tnh lin tc ti x0 trc)Tm a, b f c o hm ti x = 0

f lin tc ti x = 0 Vi b = 0:f c o hm ti x = 0 a = 2, b = 0

nh lV d:

o hm hm ngcCho y = f(x): (a, b)(c, d) lin tc v tng ngt.Khi tn ti hm ngc f 1: (c, d) (a, b) lin tc v tng ngt.Nu tn ti f (x0) 0, xo(a, b) th ti y0 = f(x0), f 1 c o hm vTa thng vit:

o hm cc hm lng gic ngc y = arcsin x, x(1, 1) x = sin y, y = arctan x, xR x = tan y,

Bng cng thc o hm cc hm mi

o hm hm cho theo tham sCho cc hm s :Nu : * x = x(t) c hm ngc t = t(x) * x(t) v y(t) c o hm, x(t) 0

V dCho :Tnh y(x) ti x =1x = 1 t.et 1 = 1 t = 0

O HM CP CAOCho f(x) c o hm cp 1 trong ln cn x0, nu f c o hm ti x0, tC th vit: Tng qut: o hm cp n l o hm ca o hm cp (n 1)

V dTm o hm cp 2 ca f ti x = 1:

o hm cp cao cc hm c bn

o hm cp cao cc hm c bn

Cng thc o hm cp cao(cng thc Leibnitz)o hm cp cao ca tng hiu:o hm cp cao ca tch:

V dTnh o hm cp 7 ti x = 1.

Tnh o hm cp 7 ti x = 1:

o hm cp cao ca hm tham sCho cc hm s :

V dCho y(t) = t2 + 1, x(t) = t3 t + 2, tnh y(x).

Cch 2:y(t) = t2 + 1, x(t) = t3 t + 2, tnh y(x).

S KH VI V VI PHNf kh vi ti x0 nu tn ti mt hng s A sao chohayKhi i lng:gi l vi phn ca f ti x0 .

Quan st

V dCho f(x) = x2 , chng minh f kh vi v tm df(1).df(1)o(x)

Quan st

o hm v vi phnf kh vi ti x0 f c o hm ti x0 .Cch vit thng thng:Cch vit khc ca o hm:

V dCho f(x) = 3x2 x, tm s gia f v vi phn df ti x = 1 vi x =0.01.

Tm vi phn ca f(x) = xex ti x = 0.

Tnh gn ng ln(1,02) nh vi phnp dng vi f(x) = lnx, x = 1.02, x0 = 1

Cnh ca khi lp phng tng ln 1cm th vi phn th tch l 12cm3 . Tm di ban u ca cnh.Gi x l cnh khi lp phng v V l th tch.Cnh tng 1cm : Vi phn th tch:

Cc php tnh vi phn

Vi phn hm hpNu y = f(x) kh vi theo x (bin c lp):Nu x = x(t) , y = f(x) kh vi, x = x(t) kh vi y = f(x(t)) kh vi theo t (bin c lp):D x l bin c lp hay hm s, dng vi phn ca y theo x khng i.

V d p dngCho y = f(x) = sin(x2), Tnh dy theo dx Vi x = x(t) = arctan(t), tnh dy theo dt ti t = 1.

Cch khc: dng vi phn hm hpTi t = 1, x = /4

VI PHN CP CAONu x l bin c lp:dx = x : l hng sNu x = x(t):dx = xdt : l hm s

V dCho y = sin(x)Tnh d2y theo dx.Nu x = ch(t), tnh d2y theo dt.

o hm hm nHm s y = f(x) xc nh bi phng trnh

()gi l hm n xc nh bi ().Cch tm y(x): ly o hm pt () theo x, gii tm y theo x v y.

Tm y(x) vi y xc nh t pt :()Ly o hm pt () theo xSo snh vi kt qu ly o hm t cc biu thcV d

V d Tm y(0) vi y = y(x) xc nh bi()Ly o hm pt () theo xT (), vi x = 0 y = 1( )Thay vo ( ):

Tm o hm ti x = 1ca hm n y = y(x) xc nh bi pt:()Ly o hm () hai theo x T (), x = 1 y = 0, thay vo () ()

Tm o hm cp 2 ti x = 1ca hm n y = y(x) xc nh bi pt:(1)(2)(3)Ly o hm (1) theo xLy o hm (2) theo x

Thay x = 1, y = 0, y = 1 vo (3)(3)

Tng kt.Tnh o hm cho 3 loi hm s (y = f(x), hm n, tham s).Nu x l bin c lp: tnh vi phn l tnh o hm Nu x = x(t) (l hm s):Vi phn cp 1 : dy = y(x)dx, sau khai trin dx theo dtVi phn cp 2: d2y = ydx2 + yd2x cui cng phi a v dt2(ch tnh n cp 2)