1. Problem: Find the derivative of the function $f(x)=3x^2-5x$ from first principles. 2. Formula: The derivative from first principles is given by $f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$. 3. Important rules: Limits are linear and you may factor and cancel common nonzero factors before taking the limit; ensure algebraic simplification is done while $h\neq 0$ then take the limit $h\to 0$. 4. Compute $f(x+h)$: $$f(x+h)=3(x+h)^2-5(x+h)=3x^2+6xh+3h^2-5x-5h$$ 5. Compute the difference $f(x+h)-f(x)$: $$f(x+h)-f(x)=6xh+3h^2-5h$$ 6. Form the difference quotient: $$\frac{f(x+h)-f(x)}{h}=\frac{6xh+3h^2-5h}{h}$$ 7. Factor $h$ from the numerator: $$\frac{h(6x+3h-5)}{h}$$ 8. Cancel the common factor $h$ (showing the cancellation): $$\frac{\cancel{h}(6x+3h-5)}{\cancel{h}}$$ 9. Simplify the quotient: $$6x+3h-5$$ 10. Take the limit as $h\to 0$: $$f'(x)=\lim_{h\to 0}(6x+3h-5)=6x-5$$ 11. Final answer: Therefore $f'(x)=6x-5$.