[tex]\bf \textit{let's make }f(x)=z\qquad thus\quad z'(x)=\cfrac{1}{x}\\\\
\textit{for the sake of disambiguiting it some}
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y=xz-x\impliedby \textit{using the product rule}
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\cfrac{dy}{dx}=[1\cdot z+x\cdot \frac{dz}{dx}]-1\implies \cfrac{dy}{dx}=z+x\cdot \cfrac{1}{x}-1\implies \cfrac{dy}{dx}=z[/tex]
