# Calculate:$y'$ for $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$ and $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+…\infty}}}}$

(1) If $y = x^{x^{x^{x^{x^{.^{.^{.^{\infty}}}}}}}}$

(2) If $y = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....\infty}}}}$

then find $y'$ in both cases

(3)If $y= \sqrt{tanx+\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+....\infty}}}}$then find $(2y-1)\frac{dy}{dx}$

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$lim_{y \rightarrow \infty}x^y=\infty$ when x>1. The ellipsis is enough to understand the nature of what you're saying. – Loki Clock Mar 21 '13 at 4:14
For the record I believe the operation to which you are referring is called a Power Tower, not sure what the other one ("(2)") is called though... – Albert Renshaw Mar 21 '13 at 8:31
@ Albert Renshaw thank you for your guidance – kalpeshmpopat Mar 21 '13 at 8:32
notation $....\infty$ looks weird for me – Cortizol Mar 21 '13 at 16:32

$y = x^{x^{x^{x^{{.^{.^{.^{}}}}}}}} = x^{y}$

Take the log of both sides.

Then $\ln y = y \ln x$

$\implies\frac{1}{y} \frac{dy}{dx} = \ln x \frac{dy}{dx} + \frac{y}{x}$

$\implies \frac{dy}{dx} = \frac{y^{2}}{x(1-y \ln x)}$ and then substitute for $y$

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## Hint

1. $$y = x^y$$
2. $$y^2 = x + y$$

As mentioned in comments, to solve the first one; you might need to take a look at some of the applications of Omega Function.

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(Note that the first requires at least tacit use of the Lambert W function.) – anon Mar 21 '13 at 3:08
@ThomasAndrews Example #3 in the applications is the same. – hjpotter92 Mar 21 '13 at 3:56
1. As $y=x^y = e^{y \log{x}}$, take derivatives of both sides:

$$y' = \left( \frac{y}{x} + y' \log{x} \right ) x^y$$

Solve for $y'$.

1. As $y^2 = x+y$, then a similar implicit differentiation as above yields

$$2 y \,y' = 1 + y'$$

Solve for $y'$.

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