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

\(f(x)=4x^{2}+5x+3\)

\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{4(x+\delta x)^2+5(x+ \delta x)+3 -(4x^{2}+5x+3)}{\delta x} \\ & = \lim_{\delta x\to \; 0} \quad \frac{4(x^2+2x\delta x +(\delta x)^2) )+5x +5 \delta x +3 -4x^2-5x-3}{\delta x}\\ & = \lim_{\delta x\to \; 0} \quad \frac{4x^2+8x\delta x +4(\delta x)^2 +5x+5\delta x +3 -4x^2 -5x -3}{\delta x}\\ & = \lim_{\delta x\to \; 0} \quad \frac {8x \delta x + 4(\delta x)^2 + 5 \delta x }{\delta x} \\ & = \lim_{\delta x\to \; 0} \quad 8x+4 \delta x +5 \\ & = \quad \quad \quad 8x +5 \end{align}