Qualification > Math

Need help!!! how to solve tis!!!

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guowie88:
Q9 onli pls help file at below  ^^
thk 4 any help

guowie88:
juz edited ^^ help me solve pls... think for 1 hour alrdy..

ashish:
y=(1-x)(1+x)-1

using product rule
let U= 1-x
du/dx = -1

V=(1+x)-1

dv/dx = -1(1+x)-2 *1

dy/dx = U dv/dx + V du/dx
     =(1-x)(-1/(1+x)2 +(-1/(1+x))
 
= (-1+x)/(1+x)2  - 1/(1+x)

after combining the 2 fraction we get
dy/dx = -2/(1+x)2


from the question y = ((1-x)/(1+x))1/2
square on both side

y2 = (1-x)(1+x)-1

diff wrt x

2y dy/dx = -2/(1+x)2

dy/dx =  -1/y(1+x)2

replace y by ((1-x)/(1+x))1/2

dy/dx =  -1/((1-x)/(1+x))1/2(1+x)2
         
= -(1+x)1/2/(1-x)1/2(1+x)2

rationalize now

dy/dx = -1/ (1+x2)1/2 (1+x)

 normal gradient = negative reciprocal of of dy/dx]

dy/dx= (1-x2)1/2 (1+x)

Deadly_king:
Good job dude :D

I would have +rep you but I need to spread the love ;)

ashish:

--- Quote from: Deadly_king on October 21, 2010, 06:22:39 am ---Good job dude :D

I would have +rep you but I need to spread the love ;)

--- End quote ---

thanks again but i am stuck on the second part , i think i didn't quite understand the question ! can you help how can normal have maximum value?

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