Find \(f'(x)\) from first principles

\(f(x)=3x^{2}-4x+1\)

\begin{align} f'(x) & = \lim_{\delta x\to \; 0} \quad \frac{f(x+\delta x)-f(x)}{\delta x}\\ & = \lim_{\delta x\to \; 0} \quad \frac{3(x+\delta x)^2-4(x+ \delta x)+1 -(3x^{2}-4x+1)}{\delta x} \\ & = \lim_{\delta x\to \; 0} \quad \frac{3(x^2+2x\delta x +(\delta x)^2) )-4x -4 \delta x +1 -3x^2+4x-1}{\delta x}\\ & = \lim_{\delta x\to \; 0} \quad \frac{3x^2+6x\delta x +3(\delta x)^2 -4x-4\delta x +1 -3x^2 +4x -1}{\delta x}\\ & = \lim_{\delta x\to \; 0} \quad \frac {6x \delta x + 3 (\delta x)^2 -4\delta x }{\delta x} \\ & = \lim_{\delta x\to \; 0} \quad 6x+3 \delta x -4 \\ & = \quad \quad \quad 6x -4 \end{align}