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The sum and difference rule for differentiable equations states:

The sum (or difference) of two differentiable functions is differentiable and [its derivative] is the sum (or difference) of their derivatives.

$$\frac{\text{d}}{\text{d}x}[f(x) + g(x)] = f'(x) + g'(x)$$ $$\frac{\text{d}}{\text{d}x}[f(x) - g(x)] = f'(x) - g'(x)$$

What is the proof of this rule?

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When I saw it, we just plugged f(x)+g(x) into the definition of derivative and out it came in a couple lines. – Ross Millikan Jun 17 '11 at 15:43
up vote 2 down vote accepted

So let $\nu(x)=f(x)+g(x)$. We have

\begin{align*} \frac{d}{dx} \bigl[\nu(x)\bigr] =\nu'(x) &= \lim_{h \to 0} \frac{\nu(x+h)-\nu(x)}{h} \\ &=\lim_{h \to 0} \frac{ f(x+h)+g(x+h) - (f(x)+g(x))}{h} \\ &= \lim_{h \to 0} \biggl[ \frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}\biggr] \\ &= f'(x) + g'(x) \end{align*}

Similarly you can do it for $f(x)-g(x)$ case. An Exercise is awaiting you, here:

Exercise. Prove that if $f$ and $g$ are differentiable, then their product $fg$ is also differentiable.

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Can this be explained intuitively, though? – Arjun Mar 5 at 0:52

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