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

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))

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

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.

To find the derivative of the given expression, we will use the chain rule along with some simplifications. Let's break down the steps:

1. Let's start by defining a function for the nested square root expression. Let f(x) = sqrt(2x + sqrt(2x + sqrt(2x))).

2. To differentiate f(x) with respect to x, we will apply the chain rule. The chain rule states that if we have a composition of functions such as f(g(x)), the derivative can be calculated as f'(g(x)) * g'(x), where f'(g(x)) represents the derivative of the outer function and g'(x) represents the derivative of the inner function.

3. Let's differentiate the outer function f(x) = sqrt(u), where u = 2x + sqrt(2x + sqrt(2x)). Applying the power rule for differentiation, we have f'(x) = (1/2) * u^(-1/2) * u'.

4. Now, we need to find the derivative of the inner function, u = 2x + sqrt(2x + sqrt(2x)). This requires another level of nested square root. Let's define g(x) = 2x + sqrt(2x + sqrt(2x)).

5. Differentiating g(x) with respect to x, we have g'(x) = 2 + (1/2) * (2x + sqrt(2x))^(-1/2) * (2 + (1/2) * (2x)^(-1/2)).

6. Finally, substituting the expressions we found for f'(x) and g'(x) into the chain rule formula, we get the derivative of the original expression:

f'(x) = (1/2) * (2x + sqrt(2x + sqrt(2x))))^(-1/2) * (2 + (1/2) * (2x + sqrt(2x))^(-1/2) * (2 + (1/2) * (2x)^(-1/2))).

So, the derivative of sqrt(2x + sqrt(2x + sqrt(2x))) would be given by f'(x) = (1/2) * (2x + sqrt(2x + sqrt(2x))))^(-1/2) * (2 + (1/2) * (2x + sqrt(2x))^(-1/2) * (2 + (1/2) * (2x)^(-1/2))).