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U kalkulusu , pravilo derivacije količnika je metoda izračunavanja derivacije funkcije koja je prikazana kao količnik druge dvije funkcije za koje derivaicja postoji.
Ako je funckija
f
(
x
)
{\displaystyle f(x)}
f
(
x
)
=
g
(
x
)
h
(
x
)
{\displaystyle f(x)={\frac {g(x)}{h(x)}}}
gdje je
h
(
x
)
{\displaystyle h(x)}
0
{\displaystyle 0}
g
(
x
)
/
h
(
x
)
{\displaystyle g(x)/h(x)}
d
d
x
f
(
x
)
=
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
h
(
x
)
2
.
{\displaystyle {\frac {d}{dx}}f(x)=f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{{h(x)}^{2}}}.}
Ili, prezicnije, za svako
x
{\displaystyle x}
otvorenom intervalu , a koje sadrži
a
{\displaystyle a}
h
(
a
)
{\displaystyle h(a)}
0
{\displaystyle 0}
g
′
(
a
)
{\displaystyle g'(a)}
h
′
(
a
)
{\displaystyle h'(a)}
f
′
(
a
)
{\displaystyle f'(a)}
f
′
(
a
)
=
g
′
(
a
)
h
(
a
)
−
g
(
a
)
h
′
(
a
)
h
(
a
)
2
.
{\displaystyle f'(a)={\frac {g'(a)h(a)-g(a)h'(a)}{h(a)^{2}}}.}
Derivacija od
(
4
x
−
2
)
/
(
x
2
+
1
)
{\displaystyle (4x-2)/(x^{2}+1)}
d
d
x
(
4
x
−
2
)
x
2
+
1
{\displaystyle {\frac {d}{dx}}{\frac {(4x-2)}{x^{2}+1}}}
=
(
x
2
+
1
)
(
4
)
−
(
4
x
−
2
)
(
2
x
)
(
x
2
+
1
)
2
{\displaystyle ={\frac {(x^{2}+1)(4)-(4x-2)(2x)}{(x^{2}+1)^{2}}}}
=
(
4
x
2
+
4
)
−
(
8
x
2
−
4
x
)
(
x
2
+
1
)
2
{\displaystyle ={\frac {(4x^{2}+4)-(8x^{2}-4x)}{(x^{2}+1)^{2}}}}
=
−
4
x
2
+
4
x
+
4
(
x
2
+
1
)
2
{\displaystyle ={\frac {-4x^{2}+4x+4}{(x^{2}+1)^{2}}}}
U gornjem primjeru, odabrali smo
g
(
x
)
=
4
x
−
2
{\displaystyle g(x)=4x-2}
h
(
x
)
=
x
2
+
1
{\displaystyle h(x)=x^{2}+1}
Analogijski, derivacija
sin
(
x
)
/
x
2
{\displaystyle \sin(x)/x^{2}}
x
{\displaystyle x}
cos
(
x
)
x
2
−
sin
(
x
)
2
x
x
4
{\displaystyle {\frac {\cos(x)x^{2}-\sin(x)2x}{x^{4}}}}
Za više informacija o derivacijama trigonometrijskih funkcija , pogledajte: Derivacija funkcije .
Drugi primjer je:
f
(
x
)
=
2
x
2
x
3
{\displaystyle f(x)={\frac {2x^{2}}{x^{3}}}}
gdje imamo
g
(
x
)
=
2
x
2
{\displaystyle g(x)=2x^{2}}
h
(
x
)
=
x
3
{\displaystyle h(x)=x^{3}}
g
′
(
x
)
=
4
x
{\displaystyle g'(x)=4x}
h
′
(
x
)
=
3
x
2
{\displaystyle h'(x)=3x^{2}}
Derivacija
f
(
x
)
{\displaystyle f(x)}
f
′
(
x
)
{\displaystyle f'(x)\,}
=
(
4
x
⋅
x
3
)
−
(
2
x
2
⋅
3
x
2
)
(
x
3
)
2
{\displaystyle ={\frac {\left(4x\cdot x^{3}\right)-\left(2x^{2}\cdot 3x^{2}\right)}{\left(x^{3}\right)^{2}}}}
=
4
x
4
−
6
x
4
x
6
{\displaystyle ={\frac {4x^{4}-6x^{4}}{x^{6}}}}
=
−
2
x
4
x
6
{\displaystyle ={\frac {-2x^{4}}{x^{6}}}}
=
−
2
x
2
{\displaystyle =-{\frac {2}{x^{2}}}}
Pretpostavimo funkciju
f
(
x
)
=
g
(
x
)
/
h
(
x
)
{\displaystyle f(x)=g(x)/h(x)}
gdje je
h
(
x
)
{\displaystyle h(x)}
g
{\displaystyle g}
h
{\displaystyle h}
f
′
(
x
)
=
lim
Δ
x
→
0
f
(
x
+
Δ
x
)
−
f
(
x
)
Δ
x
=
lim
Δ
x
→
0
g
(
x
+
Δ
x
)
h
(
x
+
Δ
x
)
−
g
(
x
)
h
(
x
)
Δ
x
{\displaystyle f'(x)=\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}=\lim _{\Delta x\to 0}{\frac {{\frac {g(x+\Delta x)}{h(x+\Delta x)}}-{\frac {g(x)}{h(x)}}}{\Delta x}}}
=
lim
Δ
x
→
0
1
Δ
x
(
g
(
x
+
Δ
x
)
h
(
x
)
−
g
(
x
)
h
(
x
+
Δ
x
)
h
(
x
)
h
(
x
+
Δ
x
)
)
{\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\left({\frac {g(x+\Delta x)h(x)-g(x)h(x+\Delta x)}{h(x)h(x+\Delta x)}}\right)}
=
lim
Δ
x
→
0
1
Δ
x
(
(
g
(
x
+
Δ
x
)
h
(
x
)
−
g
(
x
)
h
(
x
)
)
−
(
g
(
x
)
h
(
x
+
Δ
x
)
−
g
(
x
)
h
(
x
)
)
h
(
x
)
h
(
x
+
Δ
x
)
)
{\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\left({\frac {(g(x+\Delta x)h(x)-g(x)h(x))-(g(x)h(x+\Delta x)-g(x)h(x))}{h(x)h(x+\Delta x)}}\right)}
=
lim
Δ
x
→
0
1
Δ
x
(
h
(
x
)
(
g
(
x
+
Δ
x
)
−
g
(
x
)
)
−
g
(
x
)
(
h
(
x
+
Δ
x
)
−
h
(
x
)
)
h
(
x
)
h
(
x
+
Δ
x
)
)
{\displaystyle =\lim _{\Delta x\to 0}{\frac {1}{\Delta x}}\left({\frac {h(x)(g(x+\Delta x)-g(x))-g(x)(h(x+\Delta x)-h(x))}{h(x)h(x+\Delta x)}}\right)}
=
lim
Δ
x
→
0
g
(
x
+
Δ
x
)
−
g
(
x
)
Δ
x
h
(
x
)
−
g
(
x
)
h
(
x
+
Δ
x
)
−
h
(
x
)
Δ
x
h
(
x
)
h
(
x
+
Δ
x
)
{\displaystyle =\lim _{\Delta x\to 0}{\frac {{\frac {g(x+\Delta x)-g(x)}{\Delta x}}h(x)-g(x){\frac {h(x+\Delta x)-h(x)}{\Delta x}}}{h(x)h(x+\Delta x)}}}
=
lim
Δ
x
→
0
(
g
(
x
+
Δ
x
)
−
g
(
x
)
Δ
x
)
h
(
x
)
−
g
(
x
)
lim
Δ
x
→
0
(
h
(
x
+
Δ
x
)
−
h
(
x
)
Δ
x
)
h
(
x
)
h
(
lim
Δ
x
→
0
(
x
+
Δ
x
)
)
{\displaystyle ={\frac {\lim _{\Delta x\to 0}\left({\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right)h(x)-g(x)\lim _{\Delta x\to 0}\left({\frac {h(x+\Delta x)-h(x)}{\Delta x}}\right)}{h(x)h(\lim _{\Delta x\to 0}(x+\Delta x))}}}
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
[
h
(
x
)
]
2
{\displaystyle ={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}}
![{\displaystyle ={\frac {g'(x)h(x)-g(x)h'(x)}{[h(x)]^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c052044209e96b2824a60be415ce8d286e6b11c3)