Answer:
f'(x)=2
Step-by-step explanation:
[tex]if~y=ax^n\\where~a~is~constant.\\\frac{dy}{dx} =a nx^{n-1}\\f(x)=2x-1\\f'(x)=2*1x^{1-1}-0\\=2x^0\\=2*1\\=2[/tex]
