Proof of Various Derivative Properties — Question 9

Question 9

Prove from first principles that if $$f(x)=x^3,$$ then $$f'(a)=3a^2.$$

Question 9 - Solution

By definition of the derivative, $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$

Substitute $f(x)=x^3$: $$f'(a)=\lim_{h\to 0}\frac{(a+h)^3-a^3}{h}.$$

Expand the cube: $$(a+h)^3=a^3+3a^2h+3ah^2+h^3.$$

Substitute into the difference quotient: $$f'(a)=\lim_{h\to 0}\frac{a^3+3a^2h+3ah^2+h^3-a^3}{h}.$$

Simplify the numerator: $$f'(a)=\lim_{h\to 0}\frac{3a^2h+3ah^2+h^3}{h}.$$

Factor out $h$: $$f'(a)=\lim_{h\to 0}\frac{h(3a^2+3ah+h^2)}{h}.$$

Cancel $h$ (for $h\neq 0$): $$f'(a)=\lim_{h\to 0}(3a^2+3ah+h^2).$$

Now take the limit: $$f'(a)=3a^2.$$

Thus, using first principles, $$\boxed{f'(a)=3a^2}.$$