Taller No 8

  • April 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Taller No 8 as PDF for free.

More details

  • Words: 3,250
  • Pages: 22
TALLER No.8 1)

x 5 =5x 4

2) x

3)

3

=x (3 )

1

=(3)

1

1 = t3 =

2

2

( x )( 2 )

1

(

t 3 (0 ) −1 3t 2

(t ) 3

2

2

)

0 −3t 2 t6 −3t 2 = t6 =

4)

4 = u4 u 4 (0 ) −4 4u 3 = 2 u4

(

)

0 −16 u = u8 −16 u 3 = u8 5)

(

3

1 = 5u 5 5u 5 (0 ) −1 25 u 4 = 2 5u 5

( )

(

)

)

0 −25 u 4 5u 10 −25 u 4 = 5u 10 =

6)

x7 = 7 7 7x6 =

(

=

) −x ( 0 )

(7 ) 2

49 x 6 −0 49

=

49 x 6 49

7

= 1

7) 3

x

2

49 x 6 49

=

1 x

=

= =

2

x

3

2

3

[

]

(0 ) −1 2 3 ( x ) −13 (1) 2

2  x 3    

[

0 −1 2

x −2

3 x

(x)

3 4

− 1

3

]

3

( x ) −13 4

3

8) 2 x −x 3 =2 −3 x

2

9) 4 x 3 −3 x 2 +7 =12 x 2 −6 x

10) 5 −2 x 2 +x 4 =−4 x +4 x 3

11) 3 x 4 −7 x 3 +5 x 2 +8 =12 x 3 −21x 2 +10 x

12) 4 x 3 +2 +1 x x (0 ) −1(1) =12 x 2 + x2 0 −1 =12 x 2 + 2 x −1  =12 x 2 + 2  x  13) 3u 2 + 3 2

u u 2 (0 ) −3( 2u ) =6u + 2 u2 0 −6u =6u + u4

(

)

−6u  =6u + 4   u 

14) x 6 6 + 6 6 x 6(6 x 5 ) −x 6 (0 ) x 6 (0 ) −6(6 x 5 ) = + 2 6 (x 6 )2 =

36 x 5 −0 0 −36 x 5 + 36 x 12

x6 6 + 6 6 x 6(6 x 5 ) −x 6 (0 ) x 6 (0 ) −6(6 x 5 ) = + 62 (x 6 )2 36 x 5 −0 0 −36 x 5 + 36 x 12 −36 x 5  36 x 5 −0  = +  x 12  36   =

15) x 1, 2 + 1 x 0, 6 =1,2 x 0 , 2 +

x 0 , 6 (0 ) −1(0,6 x −0 , 4 )

(x ) 0,6

2

0 −0,6 x −0 , 4 =1,2 x 0 , 2 + x 1, 2 −0,6 x −0 , 4   =1,2 x 0 , 2 +   x 1, 2  

16) x 0 , 4 −x −0 , 4

=0,4 x −0 , 6 −(−0, 4 x −1, 4 )

17) 2

2

x +

=2 x 

1

x 2  + 1  x 2

2

1 − 1 =2 x 2  +2 x 2      

=2 x

1

2

+2 x

− 1

2

−1 + −x 2     1 7 7 18) x + 7 +7 x + +7 x x 7  x (0 ) −1 7 x 6 =7 x 6 + 2  x7 

=x

− 1

2

(

)

0 −7 x 6 =7 x 6 +  x 14 

−7 x 6 =7 x 6 +  x 14  19) 2

x3 +

(

) x (0 ) −7(1)  +7 +    

 0 −7   +7 + x 2    

 −7   +7 + x 2    

2 x3

3 2 =2 x 2  + 3   x 2 3 − 3 2  =2 x 2  +2 x     

=2 x

3

=3 x

1



− 3

2

+2 x

2

− 5 2  +−3 x

2

x2



2

x3 +

2

x3

3 2 =2 x 2  + 3   x 2 3 − 3 2  =2 x 2  +2 x      3

=2 x

+2 x

2

− 3

2

− 5 2  + −3 x    20) 2 t − 3 3 t 1 3 =2 t 2  − 1   t 3 1

=3 x

2

=2 t  =2t =t

21) 2 x

3

− 1

2

=3 x

22)

3

1

1

−3t

2

2

=x =x

1

3

3

=1

3

=1

3

5

3

4 1

+5 x

x −3 1

− 1

−4 3  − −t   

2

+4 x 1

− 1  3  −3 t    

2

4

1 x 1



1

x 3 − 1 −1 x 3     x x

−2

3

.−2

3

−1x

− 1

3

− −1 3 x 

−4

3

23) 3 x 4 +(2 x −1) 2

=12 x 3 +2(2 x −1)(2 ) =12 x 3 +4(2 x −1)

24)

( y −2 )(2 y −3) =( y −2 )(2 ) +(2 y −3)(1) =(2 y −4 ) +(2 y −3)

25)

( x −7 )(2 x −9 ) =( x −7 )(2 ) +(2 x −9 )(1) =(2 x −14 ) +(2 x −9 ) 2

26)  1 x +  x  1  x (0 ) −1(1)   =2x + 1 +  x x2    1 0 − 1    =2x + 1 + 2  x x   

  

2

1  x +  x 

1  x (0 ) −1(1)   =2x + 1 +  x  x2   1  0 −1   =2x + 1 + 2  x  x   1   −1  =2x + 1 + 2  x    x 

27) (u +1)(2u +1)

=(u +1)(2 ) +(2u +1)(1) =(2u +2 ) +(2u +1) 2

28)  1     x +  x  

2

1  1   1 =  x 2 + 1  x +    x    x 2

   

2

− 1  1   12  =  x +1 x 2    x +       x   −1  1  1 2  =2  x + x 2    x +    x   1 − 1 − 1 −3  =2 x 2 +1 x 2  1 2 x 2 + −1 2 x 2     

29)

(t

+1)(3t −1)

2

=(t +1)2(3t −1)(3) +(3t −1) =(t +1)6(3t +1) +(3t −1)

(u −2 )3 2 =3(u −2 ) (1) 2 =3(u −2 )

31)

( x + 2) 3 2 = 3( x + 2 ) (1) 2 = 3( x + 2 )

32)

(1)

2

=(t +1)(18 t −6 ) +(3t −1)

30)

2

2

( x +1)( x −1) 2 2 =( x +1)2( x −1)(1) +( x −1) (1) 2 =( x +1)2( x −1) +( x −1) 2 =( x +1)( 2 x −2 ) +( x −1)

   

( x +1)( x −1) 2 2 =( x +1)2( x −1)(1) +( x −1) (1) 2 =( x +1)2( x −1) +( x −1) 2 =( x +1)( 2 x −2 ) +( x −1) 3

33)  x +1     x 

 x +1   x (1) −( x +1)(1)  =3    x2  x    2

 x +1   x −( x +1)  =3    x2  x    2

3

34) 2t −1     2t 

2

2t −1  =3   2t 

2 y (2 ) −(2t −1)( 2 )      (2t ) 2  

2t −1  4t −(4t −2 )  =3    4x 2  2t    2

3

3

35)  y +2   y −2    y   +  y      

 y +2   y (1) −( y +2 )(1)   y +2   y (1) −( y −2 )(1)    =3 +3 2  y     y        y y2         2

2

 y +2   y −( y +2 )   y +2   y −( y −2 )    =3 +3 2  y     y        y y2         2

36) 2 y 2 +3 y −7 y = =

37)

y (4 y +3) −(2 y 2 +3 y −7 )(1) y2

(4 y

2

−3 y ) −(2 y 2 +3 y −7 ) y2

=

4 y 2 −3 y −2 y 2 −3 y +7 y2

=

2 y 2 +7 y2

(x

+1) x 2 x[2( x +1)(1)] −( x +1) (1) = x2 2 x[2( x +1)] −( x +1) = x2 2 2 x ( x +1) −( x +1) = x2 2

2

(x

+1) x 2 x[2( x +1)(1)] −( x +1) (1) = x2 2 x[2( x +1)] −( x +1) = x2 2 2 x ( x +1) −( x +1) = x2 2

2 38) x −3 x +1 x 2 x −3 x +1 = 1 x 2

=

=

x

1

2

(2 x −3) −(x 2

[

(x) 2

− 1

(

(x)

− 1

2

−3 x +1) 1

2

]

(1)

2

1  x 2    

x

1

2

(2 x −3) −(x 2

−3 x +1) 1

2

(1))

x 1 1 − 1 2   2 x 2 −3 x 2 −(x −3 x +1) 1 2 ( x ) 2   = x

(

)

39) t +3 t t

3 t + t = t

1

2

− 1 3 =t +  t 2    t  −3 t (0 ) −3(1) −1 2  =1 + t   2 2   t −3 0 −3 −1 2  =1 + t   2 2   t −3 −3  −1 2  =1 + 2  t   2  t 

40)

( x +1) 2 =

+( x −1) x2 2 2 x 2 [2( x +1)(1) +2( x −1)(1)] −( x +1) +( x −1) (2 x ) 2

(x ) 2

2

x 2 [2( x +1) +2( x −1)] −( x +1) +( x −1) = x4 2

(2 x )

x 2 (2 x +2 ) +(2 x −2 ) −(2 x 2 +2 x ) +( x −1) 2

=

2

4

2

( x +1) 2 =

+( x −1) x2 2 2 x 2 [2( x +1)(1) +2( x −1)(1)] −( x +1) +( x −1) (2 x ) 2

(x ) 2

2

x 2 [2( x +1) +2( x −1)] −( x +1) +( x −1) x4 2

=

2

(2 x )

x 2 (2 x +2 ) +(2 x −2 ) −(2 x 2 +2 x ) +( x −1) = x4 2

(2 x = 41)

(2t

+2 x 2 ) +(2 x −2 ) −(2 x 2 +2 x ) +( x −1) x4 2

3

+3)

2

1, 6

1,6 x 2 , 9 −2,3 x 2 , 9 =3 x 2 − x 4,6 −0,7  =3 x 2 − 4 , 6   x 

2 y +(3 y ) =2 y =y =y =y

44)

2

+(2t −3) 4t 2 2 4t [2(2t +3)(2 ) −2(2t −3)(2 )] −(2t +3) −(2t −3) (4 ) = 4t 2 2 2 4t [4(2t +3) −2(2t −3)] −(2t +3) −(2t −3) (4 ) = 4t 2 2 2 4t (8t +12 ) −(8t −12 ) −(8t +3) −(2t −3) = 4t 2 2 2 ( 32t 2 +48t ) −(8t −12 ) −(8t +3) −(2t −3) = 4t 2 2

42) x 3 − x x 2,3 x 2 , 3 (1,6 x 0 , 6 ) −x 1, 6 (2,3 x 1, 3 ) =3 x 2 − (x 2,3 )2

43)

2

1

− 1 − 1 − 1

2

2 2 2

− 1

+(3 y )

− 1

[ ] ] +[−3(3 y ) + −1(3 y )

− 2

− 2

+( −9 y )

− 2

(8 y ) 2 3 +(8 y ) −2 3 −5 − 1 =2 (8 y ) 3 (8) + −2 (8 y ) 3 (8) 3 3 =16

− 1 ( 8 y) 3 3

[

[

+ −16

(8 y ) 3

−5

3

]

]

(8 y ) =2

2

+(8 y )

3

(8 y ) 3

=16

3

− 1

3

−2

3

[

(8) + −2 3 (8 y )

[

(8 y ) −13

+ −16

3

45) (16t ) 3 4 −(16t ) −3 4

=3

(16t ) 4

=48 46)

3

(16t ) 4

2

=27t

2

=27t

2

=18t

y 47) d d x

3



− 1

4

27t

2

3

−(27t )

− 1

3

[

[

− −48

(16t ) 4

−7

−2

d x

y 49) d d u

4

4

]

]

(16 )

−2

3

3

[ 3 (27t ) (27 )] ] −[−18(27t ) − −2

=

−5

−5

3

3

y =3x 2

1 x3 x3 +

y =3x 2

+

y =x3 +

y =3x 2

u 48) d

−7

3

−1(27t )

3

]

(16 ) − −3 4 (16t )

4

3

− 1

3

]

(8)

27t 2 1

3

=27t

−5

3

1

27t 2 −

=18t

− 1

(8 y )

−5

=

=

(0 )

−1(3 x 2

(x ) 3

)

2

0 −3 x 2 x6  −3 x 2   +   x6  

5 x x ( 0 ) −5(1) u = 2 x −7 + x2 0 −5 u = 2 x −7 + x2  −5  u = 2 x −7 +  2  x 

u =x

2

−7 x +

7 +6 3u 2 3u 2 ( 0 ) −7( 6u ) y =3u 2 −1 0 u + 2 3u 2 0 −4 2 u y =3u 2 −1 0 u + 3u 4 2 u   −4 y =3u 2 −10 u +  4  3u 

y =u 3 −5u

2

+

(

)

d y d u

7 +6 3u 2 3u 2 ( 0 ) −7( 6u ) y =3u 2 −1 0 u + 2 3u 2 0 −4 2 u y =3u 2 −1 0 u + 3u 4 2 u   −4 y =3u 2 −10 u +  4  3u 

=

y =u 3 −5u

+

2

(

50) dx = dt

x =

(t

x = x = x = x = 51) y =

)

−5t 2 +7t −1 t2 3t 2 −10 t +7 − t 3 −5t 2 +7t −1 ( 2t )

3

t2

)

(

) ( (t ) +7t ) −(2t 2

(3t

4

−10 t 3

)

2

2

4

−10 t 3 +14 t 2 −2t

)

4

3t

4

−10 t

3

+7t

2

t −2t 4 +10 t 3 −14 t 2 +2t t4

t 4 −7t 2 +2t t4

x

dy  2 y  =1 dx  dy  =2 x   dx 

(

)

1 − 1   =2 x 2  1 2  x 2  (1)     

=2 x

1

2

1 x −1 2       2

1 − 1 = x 2   x 2      = x =1

52) u = 1 x 1 u = 1 x 2 − 1 u =1 x 2    

u =x

− 1

2

du  2u −3   =1  dx  −3

− 1  x 2   =1  

−3

− 1  x 2   =1  

−3

−4 −3 −3 − 1 − 1       2  2  2  − 1 x + x − 3 2 x −x 2          =1  2        

−3

 −1

− 1 2 x 2     − 1  2 x 2     − 1  2 x 2     − 1  2 x 2  

x

−3

−4

2

−3 − 1 − 1  x 2   2 x 2   + −3 x 2  

=1

u =x du  2u −3   =1  dx  −3

− 1  x 2   =1  

−3

− 1  x 2   =1  

−3

−4 −3 −3 − 1 − 1    x 2  −3 2 x 2    −1 2 x 2  + −x 2   =1        

−3

−3 −3 − 1 − 1  x 2   2 x 2   −1 2 x 2  + −3 x 2        

− 1 2 x 2     − 1  2 x 2     − 1  2 x 2     − 1  2 x 2    

−4

=1

t −3)

2

(4 )

2

(4 )

−1(3 x 2

)

)

2

   

−5(1) 2

  

−7( 6u ) 2

u

)

2

−7( 6u ) 2

)

2

u

)

t −1 ( 2t ) +14 t 2 −2t

)

−14 t 2 +2t

−3   −x 2   =1  

− 1

−4

2

 

=1

−3   −x 2   =1  

− 1

−4

2

  

=1

Related Documents

Taller No 8
April 2020 1
Taller 8
October 2019 19
Taller 8
November 2019 12
Taller 8
April 2020 7
Taller 8
April 2020 9