Upload
others
View
1
Download
0
Embed Size (px)
Citation preview
URV A7 f(x) = 2x2 + 4- r?B+? Q7 i?2 7QHHQrBM; Bb f ′(3)\
�X limh→0
2(3 + h)2 − 10
h"X lim
x→3
2x2 − 18
x − 3
*X limh→0
2(3 + h)2 + 18
h.X lim
x→3
2x2 + 18x
x − 31X lim
h→0
2h2 − 18
h − 3
UkV 6BM/ �HH p�Hm2b Q7 x 7Q` r?B+? i?2 ;`�T? Q7 f(x) =2x − 1
x + 1?�b � i�M;2Mi
HBM2 T�`�HH2H iQ i?2 HBM2 y = 3x + 1X
�X −2 �M/ 0 "X 3 �M/ 0 *X 0 �M/ 2 .X −3 �M/ 1
1X −3 �M/ 0
UjV � T�`iB+H2 Bb KQpBM; QM � bi`�B;?i HBM2 bQ i?�i Bib /BbTH�+2K2Mi 7`QK i?2Q`B;BM Bb ;Bp2M #v s(t) = t3 − 6t2 + 12t − 8X .m`BM; r?B+? Q7 i?27QHHQrBM; iBK2 BMi2`p�Hb Bb i?2 T�`iB+H2 bT22/BM; mT\
�X (2, ∞) QMHv "X (0, 3) QMHv *X (0, ∞) QMHv.X (0, 1) ∪ (2, ∞) 1X (0, 2) QMHv
U9V q?B+? Q7 i?2b2 bi�i2K2Mib �`2 h_l1\
UBV A7 f(x) = |x|- i?2M f ′(0) .L1
UBBV A7 g(x) =|x|x
- i?2M g′(0) .L1UBBBV A7 h(x) = x|x|- i?2M h′(0) = 0
�X PMHv UBV �M/ UBBV "X PMHv UBV *X PMHv UBV �M/ UBBBV.X LQM2 �`2 h`m2 1X �HH �`2 h`m2
U8V A7 i?2 MQ`K�H HBM2 iQ i?2 ;`�T? Q7 y = f(x) �i (2, 6) Bb y = 23x − 4-
r?�i Bb f ′(2)\
�X 32
"X 23
*X −32
.X 2 1X *�MMQi #2 /2i2`KBM2/
UeV :Bp2M i?2 ;`�T? Q7 y = f(x) �b b?QrM ?2`2,
0
y
x
y=f(x)
i?2M r?B+? ;`�T? #2HQr HQQFb KQbi HBF2 ;`�T? Q7 i?2 /2`Bp�iBp2 f ′(x) \
0
y yA
x x
B
00
y
x
C D y
x0
yE
0 x
UdV h?2 i�M;2Mi HBM2 iQ y =(2 +
√x)2
�i x = 1 BMi2`b2+ib i?2 y �tBbr?2`2\
�X (0, 6) "X (0, 4) *X (0, 2) .X (0, 1) 1X (0, 3)
U3V limx→0
tanπx
x secx=
�X1
π"X 0 *X 1 .X π 1X .L1
UNV 6BM/ i?2 /2`Bp�iBp2 Q7 f(x) =1
3√1 − x2
�X2x
3(1 − x2)13
"X−2x
3(1 − x2)43
*X−2x
3(1 − x2)13
.X2x
3(1 − x2)23
1X2x
3(1 − x2)43
URyV A7 y =√
x +√x- i?2M y′ =
�X1 +
√x
2√x +
√x
"X1 +
√x
4√
x +√x
*X1 + 2
√x
2√x +
√x
.X1 + 2
√x
2√x√
x +√x
1X1 + 2
√x
4√x√x +
√x
URRV :Bp2M i?2 7QHHQrBM; /�i�
f(0) = −3 f(1) = 4f ′(0) = 2 f ′(1) = 3
g(0) = 1 g(1) = 0g′(0) = −1 g′(1) = −2
-
B7 h(x) =f(g(x))
g(x)- }M/ h′(0)X
�X 7 "X 2 *X 4 .X 3 1X 1
URkVd2
dx2
(x sinx
)=
�X 2 cosx − x sinx "X x cosx + sinx *X x cosx − sinx
.X x sinx + cosx 1X x cosx − 2 sinx
URjV *QKTmi2 limx→0
sinx cosx − sinx + x sinx
x2
�X 12
"X .L1 *X 0 .X 2 1X 1
UR9Vd
dx
(sin3(x) − cos(x3)
)=
�X 3 sin2(x) cos(x) − 3x2 sin(x3)
"X 3 sin2(x) cos(x) + 3x2 sin(x3) cos(x3)
*X 3 sin2(x) cos(x) − 3x2 sin(x3) cos(x3)
.X 3 sin2(x) cos(x) + 3x2 cos(x3)
1X 3 sin2(x) cos(x) + 3x2 sin(x3)
UR8V lb2 AKTHB+Bi .Bz2`2MiB�iBQM iQ }M/ y′ B7 x2 + 5xy − 6y4 = 18
�X2x + 5y
24y3 − 5x"X
2x + 5y
24y3*X
2x
24y3 − 5x
.X2x + 5y − 18
24y31X
2x + 5y − 18
24y3 − 5x
UReV lb2 GQ;�`Bi?KB+ .Bz2`2MiB�iBQM iQ }M/ i?2 /2`Bp�iBp2 Q7 f(x) = xsinx
�X (sinx)xsinx−1 "X xsinx (cosx) (lnx)
*Xsinx
x+ (cosx) (lnx) .X xsinx
[sinx
x+ (cosx) (lnx)
]
1X x cosx + sinx
URdV A7 f(x) = xlnx- }M/ f ′(e)X
�X 0 "X 1 *X 2 .X 3 1X 4
UR3V A7 y = x2 + 2x- i?2Mdy
dx=
�X 2x + x2x−1 "X 2x + 2x *X 2x + 2x ln 2
.X (2x + 2x) ln 2 1X 2x +2x
ln 2
URNV 6BM/ f ′(1) B7 f(x) = ln
[(3x − 1)2
(x + 1)4
]
�X 1 "X −1 *X 2 .X −2 1X 5
UkyV A7 f(x) = tan−1
(2
x2
)- f ′(−1) =
�X 1 "X4
5*X −
2
5.X
2
51X −
4
5
UkRV A7 y = arcsin(x) −√1 − x2-
dy
dx=
�X1
2√1 − x2
"X1 + x
√1 − x2
*X2
√1 − x2
.Xx2
√1 − x2
1X1
√1 + x
UkkV *QKTmi2 y′′ B7 x2 + y2 = 2y + 5X
�Xd2y
dx2=
1
1 − y+
x2
(1 − y)2"X
d2y
dx2=
1
1 − y+
x2
(1 − y)3
*Xd2y
dx2=
1 + x2
(1 − y)2.X
d2y
dx2=
1 + x2
(1 − y)3
1Xd2y
dx2=
x2
1 − y−
1
(1 − y)3
UkjV A7 y = y(x) Bb /2}M2/ BKTHB+BiHv #v i?2 2[m�iBQM y ey2= 10x-
i?2Mdy
dx=
�X y + 2yey2 "X 2y2ey
2+ ey
2 *Xy
x(2y2 + 1).X
y
(2y2 + 1)
1Xy
10(2y2 + 1)
�Mbr2`b
RX " kX � jX � 9X 1 8X * eX 1 dX � 3X . NX 1RyX 1 RRX 1 RkX � RjX 1 R9X 1 R8X � ReX . RdX *R3X * RNX � kyX " kRX " kkX " kjX * k9X .