what would be the derivative of sqrt{2x+sqrt{2x+sqrt{2x}}}

Answer 1

using the chain rule, that would be
1 / 2√(2x+√(2x+√2x)) * d/dx (2x+√(2x+√2x))

[2 + (2 + 1/√2x) / 2√(2x+√2x)] / 2√(2x+√(2x+√2x))

√2x√(2x+√2x) + 2√2x + 1
------------------------------------------------------
√2x √(2x+√2x) √(2x+√(2x+√2x))

Answer 2

d/dx [ 2x + { 2x + (2x)^.5 }^.5 ]^.5
=.5 [ 2x + { 2x + (2x)^.5 }^.5 ]^-.5 * d/dx [ 2x + { 2x + (2x)^.5 }^.5 ]

=.5 [ 2x + { 2x + (2x)^.5 }^.5 ]^-.5* [ 2 + .5{ 2x + (2x)^.5 }^-.5*d/dx{ 2x + (2x)^.5 }]

= .5 [ 2x+{ 2x + (2x)^.5 }^.5 ]^-.5* [ 2 + .5{ 2x + (2x)^.5 }^-.5 *{2 +1(2x)^-.5}]
etc

Answer 3

y = √(2x + √(2x + √(2x) ) ) ----- >square both sides
= (2x + (2x + (2x)^(1/2) )^(1/2) )^(1/2)
y^2 = 2x + (2x + (2x)^(1/2) )^(1/2)
(y^2 - 2x)^2 = 2x + (2x)^(1/2) ----> square both sides again
(y^2 - 2x)^2 - 2x = (2x)^(1/2)
now differentiate with respect to x
2(y^2 - 2x)(2y dy/dx - 2) - 2 = (1/2)(2x)^(-1/2) (2)
2(y^2 - 2x)(2y dy/dx - 2) = (2x)^(-1/2) + 2
expand the left side, solve for dy/dx

Notice that for implicit derivatives there is no unique answer.
check my algebra, I did not write this out first, so there is a prob of making errors

You could of course just do a bunch of nested chain rule steps, looked formidable.