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Old 2004-08-18, 15:49   #7
mfgoode
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Jan 2004
Mumbai,India

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Cool Unusual differentiation

Quote:
Originally Posted by mfgoode
How would you differentiate

y = x^( x ) ^(x^)----------x----right up to infinity ?
In other words y =X to the power of x , to the power of x---up to infinity ?

Mally
Thank you one and all for the interest shown in this problem and the keen insight in tackling it.

The Solution: Let Y = x^x^x^X--------- to infinity.
Therefore y =x^y -------- 0 <x<=1
Taking logs logy =ylogx
Hence log y/y=logx
Hence dx/dy = [ y*(1/y) -logy*1]/ (y^2) Quotient rule y not=0

Therefore dy/dx= (y^2)/ 1-logy
Now y tends to 1 in this range as given above
Therefore dy/dx = 1 as log1 =0

This gives the slope as 1 (tan inverse 1=45*) i.e. m=1
Hence this is the eqn. of a straight line bisecting the angle (90*) in 1st. quadrant
therefore eqn of line is y=x [y=mx+c and c=0]
I remain open for further discussion on this problem

Mally
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