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Question Number 184185 by Mastermind last updated on 03/Jan/23

$$\mathrm{Differentiate},\mathrm{y}={\left({\log }_{\mathrm{e}}\mathrm{x}\right)}^{\mathrm{x}}$$$$$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by SEKRET last updated on 03/Jan/23

$$\mathbf{y}={\left(\mathbf{ln}\left(\mathbf{x}\right)\right)}^{\mathbf{x}}$$$$\mathbf{ln}\left(\mathbf{y}\right)=\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right)$$$$\frac{\mathbf{y}{}^{\prime }}{\mathbf{y}}={\mathbf{x}}^{\prime }\cdot \mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right)+\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right){}^{\prime }$$$$\mathbf{y}{}^{\prime }=\mathbf{y}\cdot (\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right)+\frac{1}{\mathbf{ln}\left(\mathbf{x}\right)})$$$$\mathbf{y}{}^{\prime }={\left(\mathbf{ln}\left(\mathbf{x}\right)\right)}^{\mathbf{x}}\cdot (\mathbf{ln}\left(\mathbf{ln}\left(\mathbf{x}\right)\right)+\frac{1}{\mathbf{ln}\left(\mathbf{x}\right)})$$$$\mathbf{ABDULAZIZ}\mathbf{ABDUVALIYEV}$$

Commented by Mastermind last updated on 03/Jan/23

$$\mathrm{Good}!$$

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