
求数学高手解题...
7个回答
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由
1-2(sinx)^2=cos2x
2sinxcosx=sin2x
推出f(x)=0.5根号3cos2x+0.5sin2x-0.5根号3
再合并下
f(x)=cos(2x+π/3)
-0.5根号3
再往下就不用我教了吧
LZ
第一问带入
第二问解方程
很基础的东西
PS:一些符号我打不出
很蛋疼
看在我这么辛苦的份上
LZ你就把分给我吧
1-2(sinx)^2=cos2x
2sinxcosx=sin2x
推出f(x)=0.5根号3cos2x+0.5sin2x-0.5根号3
再合并下
f(x)=cos(2x+π/3)
-0.5根号3
再往下就不用我教了吧
LZ
第一问带入
第二问解方程
很基础的东西
PS:一些符号我打不出
很蛋疼
看在我这么辛苦的份上
LZ你就把分给我吧
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1、解:y
=
(4
+
x)/(4
–
x),求导可得y
’
=
[1(4
–
x)
–
(4
+
x)(-1)]/(4
–
x)
2
=
8/(x
–
4)
2
,所以在点M(2,3)处切线的斜率k
=
y
’(2)
=
8/(2
–
4)
2
=
8/4
=
2,所以选
B
;
2、解:y
=
(cosx)/e
x
,求导可得y
’
=
[(-sinx)e
x
–
(cosx)e
x
]/e
2x
=
-(sinx
+
cosx)/e
x
,所以在点M(0,1)处切线的斜率k
=
y
’(0)
=
-(sin0
+
cos0)/e
0
=
-1,所以切线方程为y
=
-(x
–
0)
+
1
=
-x
+
1
=
1
–
x,所以选
B
;
3、解:y
=
x
-2/3
,求导可得y
’
=
(-2/3)x
-5/3
=
(-2/3)*
3
√(1/x
5
),所以在点M(1,1)处切线的斜率k
=
y
’(1)
=
-2/3,所以切线方程为y
=
(-2/3)(x
–
1)
+
1
=
-2x/3
+
5/3,即2x
+
3y
–
5
=
0,所以选
C
;
4、解:y
=
x
2
+
x
–
1,求导可得y
’
=
2x
+
1,所以在点(1,1)处切线的斜率k
=
y
’(1)
=
2
+
1
=
3,所以切线方程为y
=
3(x
–
1)
+
1
=
3x
–
2,即3x
–
y
–
2
=
0,所以选
B
;
5、解:把y
=
3x
+
b和y
=
x
2
+
5x
+
6联立可得x
2
+
2x
+
6
–
b
=
0
=>
根的判别式Δ=
4
–
4(6
–
b)
=
0
=>
6
–
b
=
1
=>
b
=
5,所以选
D
;
6、解:y
=
(4
+
x)/(4
–
x),参考第一题求导可得y
’
=
[1(4
–
x)
–
(4
+
x)(-1)]/(4
–
x)
2
=
8/(4
–
x)
2
,所以选
C
;
7、解:y
=
(cosx)/e
x
,参考第二题求导可得y
’
=
[(-sinx)e
x
–
(cosx)e
x
]/e
2x
=
-(sinx
+
cosx)/e
x
,所以选
A
;
8、解:y
=
e
x
cosx,求导可得y
’
=
e
x
cosx
+
e
x
(-sinx)
=
e
x
(cosx
–
sinx),所以选
A
;
9、解:
y
=
x
2
sinx,求导可得y
’
=
2xsinx
+
x
2
cosx,所以选
D
;
10、解:y
=
√(x
2
+
1),求导可得y
’
=
{(1/2)*[(x
2
+
1)]
-1/2
}*(2x)
=
[1/√(x
2
+
1)]*x
=
x/√(x
2
+
1),所以选
B
;
11、解:y
=
cos
2
x,求导可得y
’
=
(2cosx)(-sinx)
=
-2sinxcosx,所以选
C
;
12、解:y
=
ln(x
+
√(x
2
+
1)),求导可得y
’
=
{1/[x
+
√(x
2
+
1)]}*[1
+
x/√(x
2
+
1)]
=
1/√(x
2
+
1),所以选
A
;
13、解:y
=
sin(1/x),求导可得y
’
=
[cos(1/x)]*(-1)*x
-2
=
(-1/x
2
)cos(1/x),所以选
C
;
14、解:y
=
cosx
2
,求导可得y
’
=
(-sinx
2
)*(2x)
=
-2xsinx
2
,所以选
D
;
15:解:函数y
=
x
4
–
8x
2
,求导可得dy/dx
=
4x
3
–
16x,再次求导可得d
2
y/dx
2
=
12x
2
–
16,所以选
C
;
16、解:y
=
e
-2x
,求导可得y
’
=
(e
-2x
)*(-2)
=
-2e
-2x
,再次求导可得d
2
y/dx
2
=
(-2)*(e
-2x
)*(-2)
=
4e
-2x
,所以选
A
;
17、解:y
=
lnx
2
,求导可得y
’
=
(1/x
2
)*(2x)
=
2/x,再次求导可得d
2
y/dx
2
=
2*(-1)x
-2
=
-2/x
2
,所以选
B
;
18、解:等式xy
+
lnx
+
lny
=
0两边对x求导可得y
+
x(dy/dx)
+
1/x
+
(1/y)(dy/dx)
=
0
=>
(x
+
1/y)(dy/dx)
=
-(y
+
1/x)
=>
dy/dx
=
-(y
+
1/x)/(x
+
1/y)
=
-(xy
2
+
y)/(x
2
y
+
x)
=
-[y(xy
+
1)]/[x(xy
+
1)]
=
-y/x,所以选
B
。
=
(4
+
x)/(4
–
x),求导可得y
’
=
[1(4
–
x)
–
(4
+
x)(-1)]/(4
–
x)
2
=
8/(x
–
4)
2
,所以在点M(2,3)处切线的斜率k
=
y
’(2)
=
8/(2
–
4)
2
=
8/4
=
2,所以选
B
;
2、解:y
=
(cosx)/e
x
,求导可得y
’
=
[(-sinx)e
x
–
(cosx)e
x
]/e
2x
=
-(sinx
+
cosx)/e
x
,所以在点M(0,1)处切线的斜率k
=
y
’(0)
=
-(sin0
+
cos0)/e
0
=
-1,所以切线方程为y
=
-(x
–
0)
+
1
=
-x
+
1
=
1
–
x,所以选
B
;
3、解:y
=
x
-2/3
,求导可得y
’
=
(-2/3)x
-5/3
=
(-2/3)*
3
√(1/x
5
),所以在点M(1,1)处切线的斜率k
=
y
’(1)
=
-2/3,所以切线方程为y
=
(-2/3)(x
–
1)
+
1
=
-2x/3
+
5/3,即2x
+
3y
–
5
=
0,所以选
C
;
4、解:y
=
x
2
+
x
–
1,求导可得y
’
=
2x
+
1,所以在点(1,1)处切线的斜率k
=
y
’(1)
=
2
+
1
=
3,所以切线方程为y
=
3(x
–
1)
+
1
=
3x
–
2,即3x
–
y
–
2
=
0,所以选
B
;
5、解:把y
=
3x
+
b和y
=
x
2
+
5x
+
6联立可得x
2
+
2x
+
6
–
b
=
0
=>
根的判别式Δ=
4
–
4(6
–
b)
=
0
=>
6
–
b
=
1
=>
b
=
5,所以选
D
;
6、解:y
=
(4
+
x)/(4
–
x),参考第一题求导可得y
’
=
[1(4
–
x)
–
(4
+
x)(-1)]/(4
–
x)
2
=
8/(4
–
x)
2
,所以选
C
;
7、解:y
=
(cosx)/e
x
,参考第二题求导可得y
’
=
[(-sinx)e
x
–
(cosx)e
x
]/e
2x
=
-(sinx
+
cosx)/e
x
,所以选
A
;
8、解:y
=
e
x
cosx,求导可得y
’
=
e
x
cosx
+
e
x
(-sinx)
=
e
x
(cosx
–
sinx),所以选
A
;
9、解:
y
=
x
2
sinx,求导可得y
’
=
2xsinx
+
x
2
cosx,所以选
D
;
10、解:y
=
√(x
2
+
1),求导可得y
’
=
{(1/2)*[(x
2
+
1)]
-1/2
}*(2x)
=
[1/√(x
2
+
1)]*x
=
x/√(x
2
+
1),所以选
B
;
11、解:y
=
cos
2
x,求导可得y
’
=
(2cosx)(-sinx)
=
-2sinxcosx,所以选
C
;
12、解:y
=
ln(x
+
√(x
2
+
1)),求导可得y
’
=
{1/[x
+
√(x
2
+
1)]}*[1
+
x/√(x
2
+
1)]
=
1/√(x
2
+
1),所以选
A
;
13、解:y
=
sin(1/x),求导可得y
’
=
[cos(1/x)]*(-1)*x
-2
=
(-1/x
2
)cos(1/x),所以选
C
;
14、解:y
=
cosx
2
,求导可得y
’
=
(-sinx
2
)*(2x)
=
-2xsinx
2
,所以选
D
;
15:解:函数y
=
x
4
–
8x
2
,求导可得dy/dx
=
4x
3
–
16x,再次求导可得d
2
y/dx
2
=
12x
2
–
16,所以选
C
;
16、解:y
=
e
-2x
,求导可得y
’
=
(e
-2x
)*(-2)
=
-2e
-2x
,再次求导可得d
2
y/dx
2
=
(-2)*(e
-2x
)*(-2)
=
4e
-2x
,所以选
A
;
17、解:y
=
lnx
2
,求导可得y
’
=
(1/x
2
)*(2x)
=
2/x,再次求导可得d
2
y/dx
2
=
2*(-1)x
-2
=
-2/x
2
,所以选
B
;
18、解:等式xy
+
lnx
+
lny
=
0两边对x求导可得y
+
x(dy/dx)
+
1/x
+
(1/y)(dy/dx)
=
0
=>
(x
+
1/y)(dy/dx)
=
-(y
+
1/x)
=>
dy/dx
=
-(y
+
1/x)/(x
+
1/y)
=
-(xy
2
+
y)/(x
2
y
+
x)
=
-[y(xy
+
1)]/[x(xy
+
1)]
=
-y/x,所以选
B
。
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