_Posted onn September 27, 2024_ > [!theorem] **Theorem:** _Let $f$ and $g$ be functions that are [[Linearity of Differentiation|differentiable]] at the point $x$. $f/g$ is differentiable, and $\large (f/g)'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{g^2}$._ Proof: First, we note that $\large (f/g)(t) - (f/g)(x) = \frac{f(t)}{g(t)} \frac{g(x)}{g(x)} - \frac{f(x)}{g(x)}\frac{g(t)}{g(t)} = \frac{f(t)g(x) -f(x)g(t)}{g(t)g(x)}=$ $\large \frac{f(t)g(x)-f(x)g(t)+f(x)g(x)-f(x)g(x)}{g(t)g(x)} =$ $\large \frac{g(x)f(t) - g(x)f(x)-f(x)g(t)+f(x)g(x)}{g(t)g(x)} = $ $\large \frac{g(x)(f(t)-f(x) ) - f(x)(g(t)-g(x))}{g(t)g(x)}.$ By the [[Linearity of Limits]], [[Multiplicativity of Limits]], and [[Differentiability Implies Continuity|continuity of differentiable functions]], we have $\large (f/g)'(x) = \lim_{t \rightarrow x} \frac{(f/g)(t) - (f/g)(x)}{t-x} = $ $\large\lim_{t\rightarrow x} \bigg[\frac{g(x)(f(t)-f(x) ) - f(x)(g(t)-g(x))} {g(t)g(x)} \cdot \frac{1}{t-x}\bigg] =$ $\large\lim_{t\rightarrow x} \bigg[g(x)(f(t)-f(x) ) - f(x)(g(t)-g(x)) \cdot \frac{1}{g(t)g(x)} \cdot \frac{1}{t-x}\bigg] =$ $\large\lim_{t\rightarrow x} \bigg[\frac{g(x)(f(t)-f(x) ) - f(x)(g(t)-g(x))} {t-x} \cdot \frac{1}{g(t)g(x)}\bigg] =$ $\large\lim_{t\rightarrow x} \bigg[\bigg(g(x)\frac{f(t)-f(x)}{t-x} - f(x)\frac{g(t)-g(x)}{t-x}\bigg)\cdot\frac{1}{g(t)g(x)}\bigg]=$ $\large\lim_{t\rightarrow x} \bigg[g(x)\frac{f(t)-f(x)}{t-x} - f(x)\frac{g(t)-g(x)}{t-x}\bigg]\cdot\lim_{t\rightarrow x}\frac{1}{g(t)g(x)}=$ $\large g(x)\lim_{t\rightarrow x} \frac{f(t)-f(x)}{t-x} - f(x)\lim_{t-x}\frac{g(t)-g(x)}{t-x}\cdot\lim_{t\rightarrow x}\frac{1}{g(t)g(x)}=$ $\large g(x)f'(x) - f(x)g'(x)\cdot\frac{1}{[g(x)]^2}=\frac{(gf')(x)-(fg')(x)}{g^2},$the desired result. $\blacksquare$