十周理科數(shù)學(xué)答案第六周

3根據(jù)導(dǎo)數(shù)定義判斷函數(shù)f(x) 在x03f(0)

f(x)f(0)

33

x

x0

3x23所以f(x) 在x3x23f(x)|x1|x1

f(x)f(1)lim|x1|

x

x

f(x)f(1)

(x1)

x

xf(1)

f(x)f(1)

x1

x

x1xf(1f(1f(x|x1|x13根據(jù)導(dǎo)數(shù)定義判斷函數(shù)f(x) 在x03f(0)

f(x)fx

33

333函數(shù)f(x) 在x0處可導(dǎo)3(cotx)(cosx)(cosx)sinx(sinx)cosx

csc2(secx)

1

)(1)cosx(cosx)1cos2

cos2

tanxsec(cscx)

1)(1)sinx(sinx)1cos

cotx(ex)exlne

(lnx) (arccosx)

1dx

1(cos1

siny

1xyx(arctanx)

1dx

1(tan1

sec2y

1

1x11xy1(arccotx)

1dx

1(cot1

csc2y

1

1x1yxy y(2x)yx y(x3)3xyx

y(3) 3y

y(x2) x2y2 y(2x)2xln3x23x2

y(x3)16

2x31611yx35

y(x5) x5xyx

y

112

1x323yx

y(x2)2x3x2x23

y(x6)

1x6y2x33x y(2x3)(3x)56x2yx21 y(x2)(1)(1)2x1

ysinxxycosxxln

y(sinxx2)(sinx)(x2)cosxy(cosx)x(lnx)sinx1xysecx y(secx)cscxsecxsecxtanxcscxsecxcotxcscxsecxcscx(tanxcotyxlny3x

yxlnxx(lnx)lnxy(3x)arctanx3x(arctanx)3arctanx

1t1tut

utarcsintt(arcsint)arcsint ee eey

)

(ex)x

xex

(xyxx2cos y(xx2cosx)(x)(x2cosx)12xcosxx2sinx

x

11

y3xsec y(3xsecx)(3x)secx3x(secx)3xsecxln33xsecxtanx2x2 x1x

y(x21

112)

3x2121

1x2yx22 y(x22x)(x2)(2x)2x2xlny

y

)

3(1x2)3(1x2) 1

1

(1x2

(1x2yln y(lnx)(lnx)xxlnx1lnxy

y

xsinx)

x)sinx

x(sinx)

2 x2yx2e y(x2ex)(x2)exx2(ex)2xexx2exy y

sinsinxcos

)(sinx)(sinxcosx)(sinxcosx)sin(sinxcosx)

(sinxcos1cos已知函數(shù)f(x) ，并1cos

，ff(x)

1cos )1cos

(1cosx)(1cosx)(1cosx)(1cosx)(1cosx)22sin

2sinf(2

2，

f(0)

2sin (1 2yx32x24x2在點(diǎn)(1,3y3x24x4，y(1)344yx32x24x2在點(diǎn)(1,3y35(x1，y5x2．x求曲線y 在點(diǎn)(4,2)處的切線方程x 1

y 2

2，y(4) 42 1x1所以曲線y 在點(diǎn)(4,2)處的切線方程為y2x1即y x1．

(x43yx2的通過點(diǎn)(5,11131設(shè)切點(diǎn)坐標(biāo)為

x2y

30x2，0

3x

11x

2 53解得x04，所以切點(diǎn)坐標(biāo)為(4,8)，k切 x22所以過點(diǎn)(5,11y83(x4)y3x4yx33x21M(1,3切y3x26x，則 y(1)36切yx33x21M(1,3y33(x1y3xy3

1(x1y

x103f(xf(x1x32x1f3f(x)x22f(11)22 yx33x22x，直線l:ykx，且直線l與曲線相切于點(diǎn)(xy 據(jù)題意，x33x22x 3x2