Sum Rule for Derivatives

Theorem: If $$f(x)=j(x) + k(x)$$, then $$\frac{d}{dx}f(x) = \frac{d}{dx}j(x) + \frac{d}{dx}k(x)$$

Proof
By the definition of the derivative, $$\frac{d}{dx}f(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$

$$ = \lim_{h\rightarrow 0} \frac{j(x+h) + k(x+h)- (j(x) + k(x))}{h}$$.

$$ = \lim_{h\rightarrow 0} \frac{j(x+h) + k(x+h)- j(x) - k(x)}{h}$$.

$$ = \lim_{h\rightarrow 0} \frac{j(x+h) - j(x) + k(x+h) - k(x)}{h}$$.

$$ = \lim_{h\rightarrow 0} \left (\frac{j(x+h) - j(x)}{h} + \frac{k(x+h) - k(x)}{h} \right)$$

$$ = \lim_{h\rightarrow 0} \frac{j(x+h) - j(x)}{h} + \lim_{h\rightarrow 0} \frac{k(x+h) - k(x)}{h}$$.

By definition

$$= \frac{d}{dx}j(x) + \frac{d}{dx}k(x)$$

Q.E.D.