I am pretty new to Wolfram Alpha. I want Wolfram Alpha to find the derivative of $\sqrt{1+x^\sqrt{1+x^\sqrt{1+...}}}$. At the moment, I am having trouble with Wolfram Alpha simplifying it incorrectly.

I had thought that Fold[sqrt(1+{x}^#,2)&, x, Range[∞]]]

would work but Wolfram says it equals $\sqrt{x^x+1}$ which I don't think is actually equivalent.

For example, if I plug in 0 to the original equation:




$f(0)$ must be greater than $0$ because $\sqrt{1}>0$ so we know $0^{f(0)}=0$.



Now, let's plug it into the equation Wolfram Alpha gave me:





And we all know: $1\ne\sqrt{2}$

  • $\begingroup$ In Fold[sqrt(1+{x}^#,2)&, x, Range[∞]]] your brackets don't match and the ,2 seems oddly placed. If I change that to Fold[sqrt(1+x^#)&, x, Range[4]] then WA seems to understand that. WA does have a limited buffer size for input and calculations. I have seen error and omissions when it runs out of space, but I can't tell if that is what you are seeing. WA is probably isn't the tool for this task. $\endgroup$
    – Bill
    Mar 8, 2023 at 16:09
  • $\begingroup$ Note that you function can be written by an implicit function like: y= Sqrt[1+x^y] $\endgroup$ Mar 8, 2023 at 16:16
  • 1
    $\begingroup$ I'm afraid that Questions about Wolfram Alpha are off-topic in this site. $\endgroup$
    – rhermans
    Mar 8, 2023 at 16:41
  • $\begingroup$ Welcome to Mathematica StackExchange. First, as mentioned by @rhemans, this site is only about Mathematica and Wolfram Language. Second, what you observe is – in my opinion – a buggy behaviour of WolframAlpha. Writing Range[∞] is not a valid expression in Wolfram Language, and if you evaluate your Fold expression in Mathematica, you get an error. However, you also get a result where only one iteration was performed, and it seems that Wolfram|Alpha blatantly discards this error and uses the output, which is clearly wrong, as a result. $\endgroup$
    – Domen
    Mar 8, 2023 at 19:09
  • 1
    $\begingroup$ If you have access to Wolfram Mathematica, please use it, and appropriately rephrase your question, otherwise it will probably get closed ... $\endgroup$
    – Domen
    Mar 8, 2023 at 19:18

1 Answer 1


$$A = \sqrt{1 + x^{\sqrt{1 + x^{\sqrt{1+x...}}}}}$$

So $$A^2 = 1 + x^A$$

Take derivative with respect to $x$:

$$2 A(x) A^\prime (x) = x^{A(x)} \left( A(x)/x + \ln (x) A^\prime(x)\right)$$

This has not direct analytic solution for $A^\prime (x)$ (as far as I can tell).

  • 1
    $\begingroup$ In 13.1, you can calculate the implicit derivative with ImplicitD[A==Sqrt[1+x^A], A, x] to get the "result" -((A x^(-1 + A))/(-2 Sqrt[1 + x^A] + x^A Log[x])) $\endgroup$
    – Domen
    Mar 8, 2023 at 20:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.