Answer
$$
\begin{aligned} y &=4 x^{2}+3 x-2=f(x) \\ y^{\prime} &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left[4(x+h)^{2}+3(x+h)-2\right]-\left[4 x^{2}+3 x-2\right]}{h} \\ &=\lim _{h \rightarrow 0} \frac{4\left(x^{2}+2 x h+h^{2}\right)+3 x+3 h-2-4 x^{2}-3 x+2}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left.4 x^{2}+8 x h+h^{2}\right)+3 x+3 h-2-4 x^{2}-3 x+2}{h} \\ &=\lim _{h \rightarrow 0} \frac{8 x h+4 h^{2}+3 x+3 h-2-4 x^{2}-3 x+2}{h} \\ &=\lim _{h \rightarrow 0} \frac{8 x h+4 h+3)}{h} \\ &=\lim _{h \rightarrow 0}(8 x+4 h+3) \\ &=8 x+3 \end{aligned}
$$
Thus the derivative of the given function is
$$
y^{\prime}=8 x+3
$$
Work Step by Step
$$
\begin{aligned} y &=4 x^{2}+3 x-2=f(x) \\ y^{\prime} &=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left[4(x+h)^{2}+3(x+h)-2\right]-\left[4 x^{2}+3 x-2\right]}{h} \\ &=\lim _{h \rightarrow 0} \frac{4\left(x^{2}+2 x h+h^{2}\right)+3 x+3 h-2-4 x^{2}-3 x+2}{h} \\ &=\lim _{h \rightarrow 0} \frac{\left.4 x^{2}+8 x h+h^{2}\right)+3 x+3 h-2-4 x^{2}-3 x+2}{h} \\ &=\lim _{h \rightarrow 0} \frac{8 x h+4 h^{2}+3 x+3 h-2-4 x^{2}-3 x+2}{h} \\ &=\lim _{h \rightarrow 0} \frac{8 x h+4 h+3)}{h} \\ &=\lim _{h \rightarrow 0}(8 x+4 h+3) \\ &=8 x+3 \end{aligned}
$$
Thus the derivative of the given function is
$$
y^{\prime}=8 x+3
$$
