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Theorem. (Sum rule of derivatives) '
Prerequisites:
The limit definition of the derivative, $ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h} $


Proof. $ \frac{d}{dx}\bigl(f(x)+g(x)\bigr)=f'(x)+g'(x) $

From the limit definition of the derivative:

$ \begin{align} &f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &g'(x)=\lim_{h\to0}\frac{g(x+h)-g(x)}{h} \end{align} $

$ \begin{align} \frac{d}{dx}\bigl(f(x)+g(x)\bigr)&=\lim_{h\to0}\frac{f(x+h)+g(x+h)-f(x)-g(x)}{h}\\ &=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}+\lim_{h\to0}\frac{g(x+h)-g(x)}{h}\\ &=f'(x)+g'(x)\end{align} $


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