Odvod količnika

1. Osnovni primer:

   ${\displaystyle {\begin{aligned}{\frac {d}{dx}}{\frac {e^{x}}{x^{2}}}&={\frac {\left({\frac {d}{dx}}e^{x}\right)(x^{2})-(e^{x})\left({\frac {d}{dx}}x^{2}\right)}{(x^{2})^{2}}}\\&={\frac {(e^{x})(x^{2})-(e^{x})(2x)}{x^{4}}}\\&={\frac {e^{x}(x-2)}{x^{3}}}.\end{aligned}}}$ 

2. Odvod količnika se lahko uporabi za iskanje odvoda ${\displaystyle f(x)=\tan x={\tfrac {\sin x}{\cos x}}}$ :

   ${\displaystyle {\begin{aligned}{\frac {d}{dx}}\tan x&={\frac {d}{dx}}{\frac {\sin x}{\cos x}}\\&={\frac {\left({\frac {d}{dx}}\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx}}\cos x\right)}{\cos ^{2}x}}\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&={\frac {1}{\cos ^{2}x}}=\sec ^{2}x.\end{aligned}}}$ 

Naj je ${\displaystyle f(x)=g(x)/h(x).}$  Z uporabo definicije odvoda in lastnosti limite, dobimo spodnji dokaz.

${\displaystyle {\begin{aligned}f'(x)&=\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k}}\\&=\lim _{k\to 0}{\frac {{\frac {g(x+k)}{h(x+k)}}-{\frac {g(x)}{h(x)}}}{k}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k\cdot h(x)h(x+k)}}\\&=\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x+k)}{k}}\cdot \lim _{k\to 0}{\frac {1}{h(x)h(x+k)}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)+g(x)h(x)-g(x)h(x+k)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(\lim _{k\to 0}{\frac {g(x+k)h(x)-g(x)h(x)}{k}}-\lim _{k\to 0}{\frac {g(x)h(x+k)-g(x)h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&=\left(h(x)\lim _{k\to 0}{\frac {g(x+k)-g(x)}{k}}-g(x)\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k}}\right)\cdot {\frac {1}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}$ 

Naj je ${\displaystyle f(x)={\frac {g(x)}{h(x)}},}$  torej ${\displaystyle g(x)=f(x)h(x).}$  Pravilo produkta nam poda ${\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x).}$  Rešimo za ${\displaystyle f'(x)}$  in vrnemo ${\displaystyle f(x)}$  ter dobimo:

${\displaystyle {\begin{aligned}f'(x)&={\frac {g'(x)-f(x)h'(x)}{h(x)}}\\&={\frac {g'(x)-{\frac {g(x)}{h(x)}}\cdot h'(x)}{h(x)}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}$ 

Naj bo ${\displaystyle f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.}$  Tako nam pravilo kompozituma da

${\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot {\frac {d}{dx}}(h(x)^{-1}).}$ 

Da dobimo odvod v drugem delu, uporabimo odvod potence in odvod kompozituma:

${\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).}$ 

Na koncu še prepišemo kot ulomke in združimo skupne člene ter dobimo

${\displaystyle {\begin{aligned}f'(x)&={\frac {g'(x)}{h(x)}}-{\frac {g(x)h'(x)}{h(x)^{2}}}\\&={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.\end{aligned}}}$ 

Za izračun n-tega odvoda količnika lahko uporabimo implicitno odvajanje (delno v prvih n − 1 odvodih). Na primer, dvakratno odvajanje funkcije ${\displaystyle fh=g}$  (ki je ${\displaystyle f''h+2f'h'+fh''=g''}$ ) in reševanje za ${\displaystyle f''}$  poda

${\displaystyle f''=\left({\frac {g}{h}}\right)''={\frac {g''-2f'h'-fh''}{h}}.}$