_Posted on September 26, 2024_ > [!theorem] **Theorem:** _Let $f$ and $g$ be functions that are [[Linearity of Differentiation|differentiable]] at the point $x$. $fg$ is differentiable, and $(fg)'(x) = (f'g)(x) + (g'f)(x)$._ Proof: By the [[Linearity of Limits]] and [[Differentiability Implies Continuity|continuity of differentiable functions]], we have $(fg)'(x) = \lim_{t \rightarrow x} \frac{(fg)(t)-(fg)(x)}{t-x} =$ $\lim_{t \rightarrow x} \frac{(fg)(t)-(fg)(x) + (fg)(x)-(fg)(x)}{t-x} = $ $\lim_{t \rightarrow x} \frac{f(t)g(t)-f(x)g(x) + f(t)g(x)-f(t)g(x)}{t-x} = $ $\lim_{t \rightarrow x} \frac{g(x)(f(t) - f(x)) + f(t)(g(t) - g(x))}{t-x} = $ $\lim_{t\rightarrow x}g(x)\lim_{t \rightarrow x} \frac{f(t)-f(x)}{t-x} + \lim_{t\rightarrow x}f(t)\lim_{t \rightarrow x} \frac{g(t) - g(x)}{t-x} =$ $g(x)f'(x)+f(x)g'(x)= (f'g)(x) + (g'f)(x). \quad \blacksquare$