Proof of Various Derivative Properties — Question 8

Question 8

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

Question 8 - Solution

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

Substitute $f(x)=x^2$: $$f'(a)=\lim_{h\to 0}\frac{(a+h)^2-a^2}{h}.$$

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

Substitute back: $$f'(a)=\lim_{h\to 0}\frac{a^2+2ah+h^2-a^2}{h}.$$

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

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

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

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

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