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Question Number 193486 by Mingma last updated on 15/Jun/23

Answered by Subhi last updated on 15/Jun/23

$$x^x.ln\left(x\right)=ln\left(y\right)$$$$ln(x^x.ln\left(x\right))=ln\left(ln\left(y\right)\right)\Rrightarrow xln\left(x\right)+ln\left(ln\left(x\right)\right)=ln\left(ln\left(y\right)\right)$$$$ln\left(x\right)+x.\frac{1}{x}+\frac{1}{x.ln\left(x\right)}=\frac{1}{y.ln\left(y\right)}.\frac{dy}{dx}$$$$\frac{dy}{dx}=x^x.x^{x^x}.ln\left(x\right)\left(\frac{{xln}^2\left(x\right)+xln\left(x\right)+1}{xln\left(x\right)}\right)=x^{x^x+x-1}({xln}^2\left(x\right)+xln\left(x\right)+1)$$$$$$

Commented by Mingma last updated on 15/Jun/23

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