if y=x^e^-x^2,find dy/dx

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{y}={\mathrm{x}}^{{\mathrm{e}}^{-{\mathrm{x}}^2}} \\ \mathrm{Taking} \log \mathrm{on} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}, \\ \ln \left(\mathrm{y}\right)=\ln \left({\mathrm{x}}^{{\mathrm{e}}^{-{\mathrm{x}}^2}}\right) \\ \Rightarrow \ln \left(\mathrm{y}\right)={\mathrm{e}}^{-{\mathrm{x}}^2}.\ln \left(\mathrm{x}\right) \left[\mathrm{As} \ln \left({\mathrm{m}}^{\mathrm{n}}\right)=\mathrm{n}.\ln \left(\mathrm{m}\right)\right] \\ \mathrm{Differentiating} \mathrm{both} \mathrm{sides} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \mathrm{and} \mathrm{employing} \mathrm{product} \mathrm{rule}, \mathrm{we} \mathrm{get}, \\ \frac{1}{\mathrm{y}}.\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{-{\mathrm{x}}^2}.\frac{\mathrm{d}}{\mathrm{dx}}\left\{\ln \left(\mathrm{x}\right)\right\}+\ln \left(\mathrm{x}\right).\frac{\mathrm{d}}{\mathrm{dx}}\left({\mathrm{e}}^{-{\mathrm{x}}^2}\right) \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{\mathrm{x}}+\ln \left(\mathrm{x}\right)\left\{{\mathrm{e}}^{-{\mathrm{x}}^2}.\left(-2\mathrm{x}\right)\right\} \\ \Rightarrow \frac{1}{\mathrm{y}}.\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{e}}^{-{\mathrm{x}}^2}\left\{\frac{1}{\mathrm{x}}-2\mathrm{xln}\left(\mathrm{x}\right)\right\} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}.{\mathrm{e}}^{-{\mathrm{x}}^2}\left\{\frac{1-2{\mathrm{x}}^2\ln \left(\mathrm{x}\right)}{\mathrm{x}}\right\} \\ \mathbf{\Rightarrow }\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{{\mathbf{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}\mathbf{.}{\mathbf{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\left\{\frac{\mathbf{1}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{ln}\left(\mathbf{x}\right)}{\mathbf{x}}\right\}\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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