数学Paper 1讨论专区

hongji

如果不会看，你可以再calculator square他。。就发现 3/4了。。。然后你自己manual sq root => (S ...
walrein_lim88 发表于 19-7-2011 11:15 PM

    按到了，谢谢。。。

walrein_lim88

dy/dx = [e^-y] /(3 + y )
hongji 发表于 19-7-2011 11:20 PM

   你的答案也是正确的。。。y = ln ( x / ( 2 + y))
=> y = ln x  - ln (2 + y)
differentiate implicitly:
dy/dx = 1/x - 1/(2+y) (dy/dx)
(3+y)/(2+y) dy/dx = 1/x
dy/dx = (2+y)/ [x (3+y)] ....1

from y = ln x - ln (2+y)
=> y + ln (2 +y ) = ln x
=> e^(y + ln (2+y) = x
(2+y)e^(y) = x
1/x = e^(-y) / (2+y) sub into 1
dy/dx = (e^(-y)/(2+y)  x (2+y)/(3+y )     so 2 + y factor cancel left :
dy/dx = e^(-y) / (3+y)

hongji

在过程中，

1- In x /[x^2] =0
In x = 1
x = e
我想问x 怎样等于e


还有In e =1?

walrein_lim88

在过程中，

1- In x /[x^2] =0
In x = 1
x = e
我想问x 怎样等于e


还有In e =1?
hongji 发表于 19-7-2011 11:27 PM

是的。。ln e = 1 ... ln 代表 log based e ... 希望你还记得 log 5 based 5 = 1 .. log n based n = 1.. 所以 log e based e = ln e = 1

hongji

是的。。ln e = 1 ... ln 代表 log based e ... 希望你还记得 log 5 based 5 = 1 .. log n based n = 1 ...
walrein_lim88 发表于 19-7-2011 11:29 PM

    thx u , 你修economy的吧？
    可以去看economy讨论区 ?因为那边没人讨论。。谢谢

walrein_lim88

thx u , 你修economy的吧？
    可以去看economy讨论区 ?因为那边没人讨论。。谢谢
hongji 发表于 19-7-2011 11:44 PM

lol ..其实只是今天得空罢了。。之后也不懂能不能再帮忙了。。哈哈。。。好了，我去看下。。

hongji

lol ..其实只是今天得空罢了。。之后也不懂能不能再帮忙了。。哈哈。。。好了，我去看下。。
walrein_lim88 发表于 19-7-2011 11:45 PM

    了解了解。。。arigatou...

hongji

1)find the first derivative for each of the following .
(a) x [3^x]
(b) (log_10 x)^2

2)if y = x^n In x , show that  x dy/dx  = x^n  +  ny

hongji

[(1+tan x)(sec x tan x)]  -    [(sec^2x)(sec x)]  >>>        how to change to sec x(tan x-1)


i have use *sec^2x=tan^2 x +1

menglee90

回复 2688# hongji


   1.(a) use product rule:d/dx [x (3^x)] =
(d/dx x)(3^x) + (d/dx 3^x)(x)
                      =  3^x + ln 3 (3^x)(x)
                      =  3^x (1+x ln 3)


(b) d/dx
(log_10 x)^2 = d/dx
(ln x/ ln 10)^2    （先把不是base 10换去base e才能differentiate）
                                    = (1/ln10)^2   d/dx (ln x)^2
using chain rule         =  (1/ln10)^2  (2)(ln x)(1/x)
                                    = 2 ln x / [x (ln10)^2]


2. y = x^n In x
    dy/dx = x^n     d/dx (ln x) + ln x     d/dx (x^n)
              = x^n (1/x) + ln x (n)[x^(n-1)]
x dy/dx = x(x^n) (1/x) + x [x^(n-1)] (ln x) (n)
             = x^n + n ( x^n In x)
             = x^n + ny

menglee90

回复 2689# hongji

[(1+tan x)(sec x tan x)]  -    [(sec^2 x)(sec x)]
= sec x tan x + sec x tan^2 x  -   (sec^3 x)
= sec x tan x + sec x (sec^2 x -1)-   (sec^3 x)= sec x tan x + sec^3 x - sec x -   (sec^3 x)
= sec x tan x - sec x
= sec x (tan x -1)

hongji

我想问以下问题有多少的做法可以正确找到答案

find set of value x,
[x/x+1]  >=  [ 1/x+1]

我的做法是：

[x/x+1 ]    -  [1/x+1]    >=0
x-1/x+1 >=0

menglee90

回复 2692# hongji

x-1/x+1 >=0  when (x-1) and (x+1) are both positive or both negative, (x+1) not equal to 0

Case 1:
x-1>=0 and x+1>0
x>=1 and x>-1
x>=1

Case 2:
x-1<=0 and x+1<0
x<=1 and x<-1
x<-1

Combining two cases, x>=1 or x<-1

hongji

[-1 + sin x ]  / (cos ^2 x)    怎变去》》[-1]  / [1+sin x]

menglee90

回复 2694# hongji

[-1 + sin x ]  / (cos ^2 x)=   - (1 - sin x) / (1 - sin ^2 x)
= - (1 - sin x) / [(1 + sin x)(1 - sin x)]
= - 1 / (1 + sin x)

badkids91

If a+b = sqr(3sqr(3)-sqr(2)) and a-b = sqr(3sqr(2)-sqr(3)), prove that a^4+a^2b^2+b^4=4sqr(6)

badkids91

Allmaths

badkids91 发表于 23-7-2011 03:05 PM

   

Allmaths

If a+b = sqr(3sqr(3)-sqr(2)) and a-b = sqr(3sqr(2)-sqr(3)), prove that a^4+a^2b^2+b^4=4sqr(6)
badkids91 发表于 23-7-2011 02:21 PM

   

Allmaths

badkids91 发表于 23-7-2011 03:05 PM

    第5题是AM-GM。。上网查看arithmetic means-geometric means就可以学到了