How do you differentiate #d/dx sqrt(1+sqrt(1+sqrt(1+sqrt(x)))) #?

How do you differentiate
d/dx √(1+√(1+√(1+√(x)))) ?
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1 Answer
Ananda Dasgupta
May 6, 2018

#1/[32 sqrt x sqrt(1+sqrt x)sqrt (1+ sqrt(1+sqrt x))sqrt (1+ sqrt (1+ sqrt(1+sqrt x)))]#

Explanation:

To ease the writing, let's define

#y_0(x) equiv sqrt x#
#y_(n+1)(x) = sqrt(1+y_n(x))#

It is easy to see that the function we want to differentiate is #y_3(x)#

Now, using the chain rule :

#d/dx y_{n+1}(x) = 1/(2sqrt(1+y_n(x))) times d/dx(1+y_n(x))#
#qquad = 1/(2y_(n+1)(x))d/dx y_n(x)#

So,

#d/dx y_3(x) = 1/(2y_3) d/dx y_2(x) = 1/(2y_3)times 1/(2y_2) d/dx y_2(x)#
#qquad = 1/(2^4 y_3\ y_2\ y_1\ y_0)d/dx y_0(x) = 1/(32\ y_3\ y_2\ y_1\ y_0 sqrt x) #

Which can be written out as

#1/[32 sqrt x sqrt(1+sqrt x)sqrt (1+ sqrt(1+sqrt x))sqrt (1+ sqrt (1+ sqrt(1+sqrt x)))]#

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