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Question Number 184187 by Mastermind last updated on 03/Jan/23

$$\mathrm{Differentiate},\mathrm{y}={\mathrm{e}}^{\mathrm{x}}+{\mathrm{x}}^{\mathrm{x}}$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by SEKRET last updated on 03/Jan/23

$$\mathbf{y}={\mathbf{e}}^{\mathbf{x}}+{\mathbf{x}}^{\mathbf{x}}\mathbf{u}={\mathbf{e}}^{\mathbf{x}}{\mathbf{u}}^{\prime }={\mathbf{e}}^{\mathbf{x}}$$$$\mathbf{t}={\mathbf{x}}^{\mathbf{x}}\mathbf{ln}\left(\mathbf{t}\right)=\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{x}\right)$$$$\frac{\mathbf{t}{}^{\prime }}{\mathbf{t}}=\mathbf{ln}\left(\mathbf{x}\right)+1$$$$\mathbf{t}{}^{\prime }=\mathbf{t}\cdot (1+\mathbf{ln}\left(\mathbf{x}\right))={\mathbf{x}}^{\mathbf{x}}\cdot (\mathbf{ln}\left(\mathbf{x}\right)+1)$$$$\mathbf{y}{}^{\prime }=\mathbf{u}{}^{\prime }+\mathbf{t}{}^{\prime }$$$$\mathbf{y}{}^{\prime }={\mathbf{e}}^{\mathbf{x}}+{\mathbf{x}}^{\mathbf{x}}\cdot (\mathbf{ln}\left(\mathbf{x}\right)+1)$$

Commented by Mastermind last updated on 03/Jan/23

$$\mathrm{Good}!$$

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