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Question Number 75034 by chess1 last updated on 06/Dec/19

Answered by Kunal12588 last updated on 06/Dec/19

$$y=x^{x^{x^x}}$$$$logy=x^{x^x}logx$$$$log\left(logy\right)=x^{x}logx+log\left(logx\right)$$$$\frac{1}{ylogy}\frac{dy}{dx}=x^x(1+logx)logx+x^{x-1}+\frac{1}{xlogx}$$$$\frac{dy}{dx}=x^{x^{x^x}+x^x}logx[x^x(1+logx)logx+x^{x-1}+\frac{1}{xlogx}]$$

Answered by Kunal12588 last updated on 06/Dec/19

$$y=x^{x^{x^{x^x}}}$$$$logy=x^{x^{x^x}}logx$$$$\Rightarrow \frac{1}{y}\frac{dy}{dx}=x^{x^{x^x}-1}+logx\frac{d}{dx}\left(x^{x^{x^x}}\right)$$$$\Rightarrow \frac{dy}{dx}=x^{x^{x^{x^3}}}(x^{x^{x^x}-1}+logx\left(x^{x^{x^x}+x^x}logx[x^x(1+logx)logx+x^{x-1}+\frac{1}{xlogx}]\right))$$$$\Rightarrow \frac{dy}{dx}=x^{x^{x^{x^3}}}(x^{x^{x^x}-1}+x^{x^{x^x}+x^x}{\left(logx\right)}^2[x^x(1+logx)logx+x^{x-1}+\frac{1}{xlogx}])$$

Answered by Kunal12588 last updated on 06/Dec/19

$$y=x^{x^{...}}$$$$\Rightarrow y=x^y$$$$\Rightarrow \frac{dy}{dx}=x^y(\frac{y}{x}+logx\frac{dy}{dx})$$$$\Rightarrow (1-x^{y}logx)\frac{dy}{dx}=x^{y-1}y$$$$\Rightarrow \frac{dy}{dx}=\frac{\frac{x^y}{x}y}{1-ylogx}=\frac{y^2}{x(1-ylogx)}$$$$=\frac{x^{2x^{x^{....}}-1}}{(1-x^{x^{x^{...}}}logx)}$$

Commented by chess1 last updated on 06/Dec/19

$$\mathrm{thanks}$$

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