We are given the function to differentiate:
[tex]{\quad \qquad \sf \rightarrow y=x^{e^x}}[/tex]
Do take natural log on both sides, then we will be having
[tex]{:\implies \quad \sf ln(y)=ln(x^{e^{x}})}[/tex]
[tex]{:\implies \quad \sf ln(y)=e^{x}ln(x)\quad \qquad \{\because ln(a^b)=bln(a)\}}[/tex]
Now, differentiate both sides by using so called chain rule and the product rule
[tex]{:\implies \quad \sf \dfrac{1}{y}\dfrac{dy}{dx}=e^{x}ln(x)+\dfrac{e^x}{x}}[/tex]
[tex]{:\implies \quad \boxed{\bf{\dfrac{dy}{dx}=x^{e^x}\bigg\{\dfrac{e^x}{x}+ln(x)e^{x}\bigg\}}}}[/tex]
Hence, Option B) is correct
Product rule of differentiation:
- [tex]{\boxed{\bf{\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}}}}[/tex]
Where, u and v are functions of x
