> [!note] Formula
>
> If the [[Science/Math/Functions/Function|functions]] $f$ and $g$ are both [[Derivative|differentiable]], then the [[Derivative|derivative]] of their quotient can be calculated as follows:
>
> $
> \frac{d}{dx}\frac{f(x)}{g(x)} =
> \frac{
> \frac{df(x)}{dx} \cdot g(x) - f(x) \cdot \frac{dg(x)}{dx}}
> {(g(x))^2}
> $
> [!info] Proof
>
> $
> \begin{align*}
> \frac{d}{dx}\frac{f(x)}{g(x)} &=
> \lim_{h \to 0}{
> \frac{1}{h} \cdot
> \left(\frac{f(x + h)}{g(x + h)} -
> \frac{f(x)}{g(x)}
> \right)} \\
> &= \lim_{h \to 0}{
> \frac{1}{h} \cdot
> \frac{f(x + h)g(x) - f(x)g(x + h)}{g(x + h)g(x)}
> }\\
> &= \lim_{h \to 0}{
> \frac{1}{h} \cdot
> \frac{1}{g(x + h)g(x)} \cdot
> (f(x + h)g(x) - f(x)g(x + h))
> }\\
> &= \frac{1}{g(x)g(x)} \cdot \lim_{h \to 0}{
> \frac{1}{h} \cdot
> (f(x + h)g(x) - f(x)g(x + h))
> }\\
> &= \frac{1}{(g(x))^2} \cdot \lim_{h \to 0}{
> \frac{1}{h} \cdot
> (f(x + h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x + h))
> }\\
> &= \frac{1}{(g(x))^2} \cdot \lim_{h \to 0}{
> \frac{1}{h} \cdot
> ((f(x + h) - f(x))g(x) - f(x)(g(x + h) - g(x)))
> }\\
> &= \frac{1}{(g(x))^2} \cdot \left(\lim_{h \to 0}{
> \frac{(f(x + h) - f(x))g(x)}{h}
> } - \lim_{h \to 0}{
> \frac{f(x)(g(x + h) - g(x))}{h}
> }\right) \\
> &= \frac{1}{(g(x))^2} \cdot \left(\lim_{h \to 0}{
> \frac{(f(x + h) - f(x))g(x)}{h}
> } - \lim_{h \to 0}{
> \frac{f(x)(g(x + h) - g(x))}{h}
> }\right) \\
> &= \frac{1}{(g(x))^2} \cdot \left(g(x) \cdot \lim_{h \to 0}{
> \frac{f(x + h) - f(x)}{h}
> } - f(x) \cdot \lim_{h \to 0}{
> \frac{g(x + h) - g(x)}{h}
> }\right) \\
> &= \frac{1}{(g(x))^2} \cdot \left(
> \frac{df(x)}{dx} \cdot g(x) - f(x) \cdot \frac{dg(x)}{dx} \right) \\
> &= \frac{
> \frac{df(x)}{dx} \cdot g(x) - f(x) \cdot \frac{dg(x)}{dx}}
> {(g(x))^2}
> \end{align*}
> $