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Question Number 184188 by Mastermind last updated on 03/Jan/23

$$\mathrm{Differentiate},\mathrm{y}={\mathrm{x}}^{\mathrm{x}-1}$$$$$$$$$$$$\mathrm{hi}$$

Answered by SEKRET last updated on 03/Jan/23

$$\mathbf{y}={\mathbf{x}}^{\mathbf{x}-1}$$$$\mathbf{ln}\left(\mathbf{y}\right)=\mathbf{ln}\left({\mathbf{x}}^{\mathbf{x}-1}\right)$$$$\mathbf{ln}\left(\mathbf{y}\right)=(\mathbf{x}-1)\cdot \mathbf{ln}\left(\mathbf{x}\right)$$$$\mathbf{ln}\left(\mathbf{y}\right)=\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{x}\right)-\mathbf{ln}\left(\mathbf{x}\right)$$$$\left(\mathbf{ln}\left(\mathbf{y}\right)\right){}^{\prime }=(\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{x}\right)-\mathbf{ln}\left(\mathbf{x}\right)){}^{\prime }$$$$\frac{\mathbf{y}{}^{\prime }}{\mathbf{y}}=\mathbf{ln}\left(\mathbf{x}\right)+1-\frac{1}{\mathbf{x}}$$$${\mathbf{y}}^{\prime }=\mathbf{y}\cdot (\mathbf{ln}\left(\mathbf{x}\right)+1-\frac{1}{\mathbf{x}})$$$$\mathbf{y}{}^{\prime }={\mathbf{x}}^{\mathbf{x}-1}\cdot (\mathbf{ln}\left(\mathbf{x}\right)+1-\frac{1}{\mathbf{x}})$$$$\mathbf{y}{}^{\prime }={\mathbf{x}}^{\mathbf{x}-2}\cdot (\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{x}\right)+\mathbf{x}-1)$$

Commented by Mastermind last updated on 03/Jan/23

$$\mathrm{Good}!$$

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