אם f , g {\displaystyle f,g} גזירות ומתקיים g ′ ≠ 0 {\displaystyle g'\neq 0} , אזי f g {\displaystyle {\frac {f}{g}}} גזירה ומתקיים d d x ( f ( x ) g ( x ) ) = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 {\displaystyle {\frac {d}{dx}}\left({\frac {f(x)}{g(x)}}\right)={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}} .
◼ {\displaystyle \blacksquare }
נגדיר h ( x ) = f ( x ) g ( x ) {\displaystyle h(x)={\frac {f(x)}{g(x)}}} ומכאן f ( x ) = g ( x ) h ( x ) {\displaystyle f(x)=g(x)h(x)} . נשתמש בכלל למכפלת נגזרות ונקבל:
![{\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {{\dfrac {f(x+h)}{g(x+h)}}-{\dfrac {f(x)}{g(x)}}}{h}}&=\lim _{h\to 0}{\dfrac {\dfrac {f(x+h)g(x)-f(x)g(x+h)}{h}}{g(x)g(x+h)}}\\&=\lim _{h\to 0}{\dfrac {\dfrac {f(x+h)g(x)-{\color {blue}f(x)g(x)}-f(x)g(x+h)+{\color {blue}f(x)g(x)}}{h}}{g(x)g(x+h)}}\\&=\lim _{h\to 0}{\dfrac {g(x)\cdot {\dfrac {f(x+h)-f(x)}{h}}-f(x)\cdot {\dfrac {g(x+h)-g(x)}{h}}}{g(x)g(x+h)}}\\&={\dfrac {\lim \limits _{h\to 0}\left[g(x)\cdot {\dfrac {f(x+h)-f(x)}{h}}-f(x)\cdot {\dfrac {g(x+h)-g(x)}{h}}\right]}{\lim \limits _{h\to 0}g(x)g(x+h)}}\\&={\dfrac {\lim \limits _{h\to 0}g(x)\cdot {\dfrac {f(x+h)-f(x)}{h}}-\lim \limits _{h\to 0}f(x)\cdot {\dfrac {g(x+h)-g(x)}{h}}}{\lim \limits _{h\to 0}g(x)g(x+h)}}\\&={\dfrac {\lim \limits _{h\to 0}g(x)\cdot \lim \limits _{h\to 0}{\dfrac {f(x+h)-f(x)}{h}}-\lim \limits _{h\to 0}f(x)\cdot \lim \limits _{h\to 0}{\dfrac {g(x+h)-g(x)}{h}}}{\lim \limits _{h\to 0}g(x)\cdot \lim \limits _{h\to 0}g(x+h)}}\\&={\dfrac {g(x)\cdot \lim \limits _{h\to 0}{\dfrac {f(x+h)-f(x)}{h}}-f(x)\cdot \lim \limits _{h\to 0}{\dfrac {g(x+h)-g(x)}{h}}}{g(x)^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ece43052637fa8c6bde651eadf91dca8fd4443a0)
