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Question Number 184184 by Mastermind last updated on 03/Jan/23

$$\mathrm{y}={\left(\mathrm{sinx}\right)}^{\mathrm{x}}$$$$$$$$\mathrm{Differentiate}$$

Answered by SEKRET last updated on 03/Jan/23

$$\mathbf{y}={\left(\mathbf{sin}\left(\mathbf{x}\right)\right)}^{\mathbf{x}}$$$$\mathbf{ln}\left(\mathbf{y}\right)=\mathbf{x}\cdot \mathbf{ln}\left(\mathbf{sin}\left(\mathbf{x}\right)\right)$$$$\frac{{\mathbf{y}}^{\prime }}{\mathbf{y}}=\mathbf{ln}\left(\mathbf{sin}\left(\mathbf{x}\right)\right)+\mathbf{x}\cdot \frac{\mathbf{cos}\left(\mathbf{x}\right)}{\mathbf{sin}\left(\mathbf{x}\right)}$$$$\mathbf{y}{}^{\prime }={\left(\mathbf{sin}\left(\mathbf{x}\right)\right)}^{\mathbf{x}}\cdot (\mathbf{ln}\left(\mathbf{sin}\left(\mathbf{x}\right)\right)+\mathbf{x}\cdot \mathbf{ctg}\left(\mathbf{x}\right))$$

Commented by Mastermind last updated on 03/Jan/23

$$\mathrm{V}.\mathrm{good}\mathrm{of}\mathrm{you},\mathrm{everything}\mathrm{is}\mathrm{correct}$$$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{for}\mathrm{answering}\mathrm{it}$$

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