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If $f$ and $g$ are functions that are differentiable at $x$, then the derivatives of their sum and difference both exist and are given by
$$\frac{d}{dx} \left[ f\,(x) \pm g(x) \right] = f\,'(x) \pm g'(x)$$Proof:
$$\begin{array}{rcl} \displaystyle{\frac{d}{dx} \left[ f\,(x) \pm g(x) \right]} & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{ [f\,(x+h) \pm g(x+h)] - [f\,(x) \pm g(x)]}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{[f\,(x+h) - f\,(x)] \pm [g(x+h) - g(x)]}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \left[ \frac{f\,(x+h) - f\,(x)}{h} \pm \frac{g(x+h) - g(x)}{h} \right]}\\\\ & = &\displaystyle{ \left[ \lim_{h \rightarrow 0} \ \frac{f\,(x+h) - f\,(x)}{h} \right] \pm \left[ \lim_{h \rightarrow 0} \ \frac{g(x+h) - g(x)}{h} \right]}\\\\ & = & \displaystyle{f\,'(x) \pm g'(x)} \end{array}$$
QED.