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If $f$ and $g$ are functions that are differentiable at $x$, then the derivative of their product exists and is given by
$$\frac{d}{dx} \left[ f\,(x) g(x) \right] = f\,(x) g'(x) + f\,'(x) g(x)$$Proof:
After appealing to the limit definition of the derivative, we add a well-chosen value of zero, $f\,(x+h)g(x) - f\,(x+h)g(x)$, to the numerator: \(\require{color}\)
$$\begin{array}{rcl} \dfrac{d}{dx} \left[ f\,(x) g(x) \right] & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{ f\,(x+h) g(x+h) - f\,(x) g(x)}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{ f\,(x+h) g(x+h) \, {\color{red}{ - \, f\,(x+h) g(x) + f\,(x+h) g(x)}} - f\,(x) g(x)}{h}} \end{array}$$We then manipulate this expression using a bit of algebra and the limit laws to reveal the same limits that appear in the definitions of $f\,'(x)$ and $g'(x)$.
$$\begin{array}{rcl} \dfrac{d}{dx} \left[ f\,(x) g(x) \right] & = & \displaystyle{\lim_{h \rightarrow 0} \ \frac{ f\,(x+h) [g(x+h) - g(x)] + [f\,(x+h) - f\,(x)] g(x)}{h}}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ f\,(x+h) \cdot \frac{ g(x+h) - g(x) }{h} + \lim_{h \rightarrow 0} \ \frac{f\,(x+h) - f\,(x)}{h} \cdot g(x)}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \ f\,(x+h) \cdot {\color{red}{\lim_{h \rightarrow 0} \ \frac{ g(x+h) - g(x) }{h}}} \, + \, {\color{red}{\lim_{h \rightarrow 0} \ \frac{f\,(x+h) - f\,(x)}{h}}} \cdot \lim_{h \rightarrow 0} \ g(x)} \end{array}$$Replacing the limits in red above with $f\,'(x)$ and $g'(x)$ respectively, and noting the last limit is of an expression that does not depend on $h$ (effectively the limit of a constant), we have
$$\frac{d}{dx} \left[ f\,(x) g(x) \right] = \lim_{h \rightarrow 0} \ f\,(x+h) \cdot g'(x) + f\,'(x) g(x)$$To simplify the remaining limit, we note that if $f$ is differentiable at $x$, it must also be continuous there. As such,
$$\lim_{h \rightarrow 0} \ f\,(x+h) = \lim_{u \rightarrow x} \ f\,(u) = f(x)$$Making this replacement, we arrive at our desired result:
$$\frac{d}{dx} \left[ f\,(x) g(x) \right] = f\,(x) g'(x) + f\,'(x) g(x)$$ QED.