Bonjour,
[tex]Soit\;G(x)=x\ln(x)-x\\\\G'(x)=\dfrac{d}{dx}\left(x\ln(x)\right)-\dfrac{d}{dx}(x)\\\\\dfrac{d}{dx}(x\ln(x))-1=(u\times v)'+1=u'v+v'u-1\\\\u=x\;\quad v=\ln(x)\\u'=1\;\quad v'=\dfrac{1}{x}\\\\\dfrac{d}{dx}(x\ln(x))-\dfrac{d}{dx}(x)=\ln(x)+x\times\dfrac{1}{x}-1=\ln(x)+1-1=\ln(x)[/tex]
