## Calculus: Early Transcendentals 8th Edition

$$f'(x)=\ln(x)$$
$f(x)=x\ln(x)-x$ We should use the product rule that says: For $f(x)=u(x)v(x)-->f'(x)=u'(x)v(x)+u(x)v'(x)$ $f'(x)=[\frac{d}{dx}(x)]*[\ln(x)]+[x]*[\frac{d}{dx}\ln(x)]-1$ $f'(x)=[(1)(\ln(x))+(x)(\frac{1}{x})]-1$ $f'(x)=\ln(x)+1-1$ $$f'(x)=\ln(x)$$