Proof of Various Derivative Properties — Question 1

Question 1

Assume that $f$ and $g$ are differentiable at $x=a$. Prove that $$(f+g)'(a)=f'(a)+g'(a).$$

Question 1 - Solution

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

Expand the numerator: $$(f+g)(a+h)-(f+g)(a) = \bigl(f(a+h)+g(a+h)\bigr)-\bigl(f(a)+g(a)\bigr).$$

Group terms: $$= \bigl(f(a+h)-f(a)\bigr)+\bigl(g(a+h)-g(a)\bigr).$$

Substitute back into the limit: $$(f+g)'(a) = \lim_{h\to 0} \left[ \frac{f(a+h)-f(a)}{h} + \frac{g(a+h)-g(a)}{h} \right].$$

Use the limit law for sums: $$(f+g)'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} + \lim_{h\to 0}\frac{g(a+h)-g(a)}{h}.$$

Since $f$ and $g$ are differentiable at $a$, both limits exist and equal $f'(a)$ and $g'(a)$, respectively.

Therefore, $$(f+g)'(a)=f'(a)+g'(a).$$