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# Calculus. Is this correct? Because I just discovered it.

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Find the derivative of $$x^{x^{\cdot\cdot\cdot}}$$

This are infinitely many 'x's and seems impossible, but I found something really awesome.

Let $$f(x) = x^{f(x)}$$ so that the solution of $$f(x)$$ is $$x^{x^{\cdot\cdot\cdot}}$$.

$$f(x) = x^{f(x)}\\ \ln f(x) = f(x) \cdot \ln x\\ \dfrac{1}{f(x)}\cdot\dfrac{df(x)}{dx} = \dfrac{df(x)}{dx} \cdot \ln x + \dfrac{f(x)}{x}\\ \dfrac{1}{f(x)}\cdot\dfrac{df(x)}{dx} - \dfrac{df(x)}{dx} \cdot \ln x = \dfrac{f(x)}{x}\\ \dfrac{df(x)}{dx}\left(\dfrac{1}{f(x)}-\ln x\right) = \dfrac{f(x)}{x}\\ \dfrac{df(x)}{dx} =\dfrac{f(x)}{x\left({\dfrac{1}{f(x)}-\ln x}\right)}\\ \dfrac{df(x)}{dx} = \dfrac{\left(f(x)\right)^2}{x - x\cdot f(x)\cdot \ln x}\\ \dfrac{df(x)}{dx}=\dfrac{\left(x^{x^{\cdot\cdot\cdot}}\right)^2}{x - x^{x^{x^{\cdot\cdot\cdot}}+1}\cdot \ln x}$$

Jan 30, 2017

### 2+0 Answers

#1
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yes its right

Jan 30, 2017
#2
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Am I the first one to discover this solution?

MaxWong  Jan 30, 2017