1. Problem statement: We need the derivative of $f(x)=3x^{2}+2x$ using the first principle. 2. Formula: The derivative by first principle is defined by $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ 3. Compute $f(x+h)$ by substituting $x+h$ for $x$. $$f(x+h)=3(x+h)^{2}+2(x+h)$$ 4. Expand and simplify $f(x+h)$. $$f(x+h)=3(x^{2}+2xh+h^{2})+2x+2h$$ $$f(x+h)=3x^{2}+6xh+3h^{2}+2x+2h$$ 5. Form the difference $f(x+h)-f(x)$ and simplify. $$f(x+h)-f(x)=(3x^{2}+6xh+3h^{2}+2x+2h)-(3x^{2}+2x)$$ $$=6xh+3h^{2}+2h$$ 6. Divide the difference by $h$ as required by the definition. $$\frac{f(x+h)-f(x)}{h}=\frac{6xh+3h^{2}+2h}{h}$$ 7. Factor out $h$ from the numerator to cancel with the denominator. $$=\frac{h(6x+3h+2)}{h}$$ 8. Show the cancellation explicitly. $$=\frac{\cancel{h}(6x+3h+2)}{\cancel{h}}$$ 9. After canceling $h$, simplify the expression. $$=6x+3h+2$$ 10. Take the limit as $h\to0$ to find the derivative. $$f'(x)=\lim_{h\to0}(6x+3h+2)$$ $$=6x+2$$ 11. Final answer: The derivative by first principle is $f'(x)=6x+2$.