Answer:
[tex]f'(x)=1+\ln x[/tex]
Step-by-step explanation:
[tex]f'(x)[/tex] is notation for the first derivative of [tex]f(x)[/tex].
Recall the product rule:
[tex](f\cdot g)'=f'\cdot g+g'\cdot f[/tex]
Therefore, we have:
[tex]\displaystyle \frac{d}{dx}(x\ln x)=\frac{d}{dx}(x)\cdot \ln(x)+\frac{d}{dx}(\ln x)\cdot x[/tex]
Note that:
[tex]\displaystyle \frac{d}{dx}(x)=1,\\\\\frac{d}{dx}(\ln x)=\frac{1}{x}[/tex]
Simplify:
[tex]\displaystyle \frac{d}{dx}(x\ln x)=1\cdot \ln x+\frac{1}{x}\cdot x, \\\\\frac{d}{dx}(x\ln x)= \ln x+1=\boxed{1+\ln x}[/tex]
