Find the derivative of the function f(x)=2x^2-x+1 from first principles.

f(x) =  2x^2-x+ 1

The derivative of f(x) is defined as :

d/dx (f(x) )= f'(x) =  Limit  f(x+deltax))-f(x).)/deltax  as deltax tends to zero. Call deltax = h for convenience of writing.

Then d/dx(f(x)) = d/dx ({2x^2-x+1}) = f'(x) = limt[f(x+h-f(x)]/h as h -->0

=Limit{[2(x+h)^2-(x+h)+1]-[2x^2-x+1]}/h as h-->0

=Limit{2(x^2+2hx+h^2 -(x+h)+1 )-2x^2 +x -1}h as h-->

=Limit{2x^2+4hx+h^2-x-h-2x^2+x-1}/h as h-->0

=Limit{0*x^2+4hx-h+h^2}/h as h-->0

=Limit {4hx/h -h/h+h} as h-->0

=Limit 4x-1 +h as h-->0

=4x-1+0

=4x -1.

Therefore, f'(x) = d/dx(2x^2-x+1) = 4x-1 derived from first principles.