Formula for the derivative of a ratio of functions
In calculus , the quotient rule is a method of finding the derivative of a function that is the ratio of two differentiable functions.[1] [2] [3] Let
h
(
x
)
=
f
(
x
)
g
(
x
)
{\displaystyle h(x)={\frac {f(x)}{g(x)))}
, where both f and g are differentiable and
g
(
x
)
≠
0.
{\displaystyle g(x)\neq 0.}
The quotient rule states that the derivative of h (x ) is
h
′
(
x
)
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
(
g
(
x
)
)
2
.
{\displaystyle h'(x)={\frac {f'(x)g(x)-f(x)g'(x)}{(g(x))^{2))}.}
It is provable in many ways by using other derivative rules .
Reciprocal rule
The reciprocal rule is a special case of the quotient rule in which the numerator
f
(
x
)
=
1
{\displaystyle f(x)=1}
. Applying the quotient rule gives
h
′
(
x
)
=
d
d
x
[
1
g
(
x
)
]
=
0
⋅
g
(
x
)
−
1
⋅
g
′
(
x
)
g
(
x
)
2
=
−
g
′
(
x
)
g
(
x
)
2
.
{\displaystyle h'(x)={\frac {d}{dx))\left[{\frac {1}{g(x)))\right]={\frac {0\cdot g(x)-1\cdot g'(x)}{g(x)^{2))}={\frac {-g'(x)}{g(x)^{2))}.}
Utilizing the chain rule yields the same result.
Proofs
Proof from derivative definition and limit properties
Let
h
(
x
)
=
f
(
x
)
g
(
x
)
.
{\displaystyle h(x)={\frac {f(x)}{g(x))).}
Applying the definition of the derivative and properties of limits gives the following proof, with the term
f
(
x
)
g
(
x
)
{\displaystyle f(x)g(x)}
added and subtracted to allow splitting and factoring in subsequent steps without affecting the value:
h
′
(
x
)
=
lim
k
→
0
h
(
x
+
k
)
−
h
(
x
)
k
=
lim
k
→
0
f
(
x
+
k
)
g
(
x
+
k
)
−
f
(
x
)
g
(
x
)
k
=
lim
k
→
0
f
(
x
+
k
)
g
(
x
)
−
f
(
x
)
g
(
x
+
k
)
k
⋅
g
(
x
)
g
(
x
+
k
)
=
lim
k
→
0
f
(
x
+
k
)
g
(
x
)
−
f
(
x
)
g
(
x
+
k
)
k
⋅
lim
k
→
0
1
g
(
x
)
g
(
x
+
k
)
=
lim
k
→
0
[
f
(
x
+
k
)
g
(
x
)
−
f
(
x
)
g
(
x
)
+
f
(
x
)
g
(
x
)
−
f
(
x
)
g
(
x
+
k
)
k
]
⋅
1
g
(
x
)
2
=
[
lim
k
→
0
f
(
x
+
k
)
g
(
x
)
−
f
(
x
)
g
(
x
)
k
−
lim
k
→
0
f
(
x
)
g
(
x
+
k
)
−
f
(
x
)
g
(
x
)
k
]
⋅
1
g
(
x
)
2
=
[
lim
k
→
0
f
(
x
+
k
)
−
f
(
x
)
k
⋅
g
(
x
)
−
f
(
x
)
⋅
lim
k
→
0
g
(
x
+
k
)
−
g
(
x
)
k
]
⋅
1
g
(
x
)
2
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
.
{\displaystyle {\begin{aligned}h'(x)&=\lim _{k\to 0}{\frac {h(x+k)-h(x)}{k))\\&=\lim _{k\to 0}{\frac ((\frac {f(x+k)}{g(x+k)))-{\frac {f(x)}{g(x)))}{k))\\&=\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x+k)}{k\cdot g(x)g(x+k)))\\&=\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x+k)}{k))\cdot \lim _{k\to 0}{\frac {1}{g(x)g(x+k)))\\&=\lim _{k\to 0}\left[{\frac {f(x+k)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+k)}{k))\right]\cdot {\frac {1}{g(x)^{2))}\\&=\left[\lim _{k\to 0}{\frac {f(x+k)g(x)-f(x)g(x)}{k))-\lim _{k\to 0}{\frac {f(x)g(x+k)-f(x)g(x)}{k))\right]\cdot {\frac {1}{g(x)^{2))}\\&=\left[\lim _{k\to 0}{\frac {f(x+k)-f(x)}{k))\cdot g(x)-f(x)\cdot \lim _{k\to 0}{\frac {g(x+k)-g(x)}{k))\right]\cdot {\frac {1}{g(x)^{2))}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2))}.\end{aligned))}
The limit evaluation
lim
k
→
0
1
g
(
x
+
k
)
g
(
x
)
=
1
g
(
x
)
2
{\displaystyle \lim _{k\to 0}{\frac {1}{g(x+k)g(x)))={\frac {1}{g(x)^{2))))
is justified by the differentiability of
g
(
x
)
{\displaystyle g(x)}
, implying continuity, which can be expressed as
lim
k
→
0
g
(
x
+
k
)
=
g
(
x
)
{\displaystyle \lim _{k\to 0}g(x+k)=g(x)}
.
Proof using implicit differentiation
Let
h
(
x
)
=
f
(
x
)
g
(
x
)
,
{\displaystyle h(x)={\frac {f(x)}{g(x))),}
so that
f
(
x
)
=
g
(
x
)
h
(
x
)
.
{\displaystyle f(x)=g(x)h(x).}
The product rule then gives
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
+
g
(
x
)
h
′
(
x
)
.
{\displaystyle f'(x)=g'(x)h(x)+g(x)h'(x).}
Solving for
h
′
(
x
)
{\displaystyle h'(x)}
and substituting back for
h
(
x
)
{\displaystyle h(x)}
gives:
h
′
(
x
)
=
f
′
(
x
)
−
g
′
(
x
)
h
(
x
)
g
(
x
)
=
f
′
(
x
)
−
g
′
(
x
)
⋅
f
(
x
)
g
(
x
)
g
(
x
)
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
.
{\displaystyle {\begin{aligned}h'(x)&={\frac {f'(x)-g'(x)h(x)}{g(x)))\\&={\frac {f'(x)-g'(x)\cdot {\frac {f(x)}{g(x)))}{g(x)))\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2))}.\end{aligned))}
Proof using the reciprocal rule or chain rule
Let
h
(
x
)
=
f
(
x
)
g
(
x
)
=
f
(
x
)
⋅
1
g
(
x
)
.
{\displaystyle h(x)={\frac {f(x)}{g(x)))=f(x)\cdot {\frac {1}{g(x))).}
Then the product rule gives
h
′
(
x
)
=
f
′
(
x
)
⋅
1
g
(
x
)
+
f
(
x
)
⋅
d
d
x
[
1
g
(
x
)
]
.
{\displaystyle h'(x)=f'(x)\cdot {\frac {1}{g(x)))+f(x)\cdot {\frac {d}{dx))\left[{\frac {1}{g(x)))\right].}
To evaluate the derivative in the second term, apply the reciprocal rule , or the power rule along with the chain rule :
d
d
x
[
1
g
(
x
)
]
=
−
1
g
(
x
)
2
⋅
g
′
(
x
)
=
−
g
′
(
x
)
g
(
x
)
2
.
{\displaystyle {\frac {d}{dx))\left[{\frac {1}{g(x)))\right]=-{\frac {1}{g(x)^{2))}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2))}.}
Substituting the result into the expression gives
h
′
(
x
)
=
f
′
(
x
)
⋅
1
g
(
x
)
+
f
(
x
)
⋅
[
−
g
′
(
x
)
g
(
x
)
2
]
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
=
g
(
x
)
g
(
x
)
⋅
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
.
{\displaystyle {\begin{aligned}h'(x)&=f'(x)\cdot {\frac {1}{g(x)))+f(x)\cdot \left[{\frac {-g'(x)}{g(x)^{2))}\right]\\&={\frac {f'(x)}{g(x)))-{\frac {f(x)g'(x)}{g(x)^{2))}\\&={\frac {g(x)}{g(x)))\cdot {\frac {f'(x)}{g(x)))-{\frac {f(x)g'(x)}{g(x)^{2))}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2))}.\end{aligned))}
Proof by logarithmic differentiation
Let
h
(
x
)
=
f
(
x
)
g
(
x
)
.
{\displaystyle h(x)={\frac {f(x)}{g(x))).}
Taking the absolute value and natural logarithm of both sides of the equation gives
ln
|
h
(
x
)
|
=
ln
|
f
(
x
)
g
(
x
)
|
{\displaystyle \ln |h(x)|=\ln \left|{\frac {f(x)}{g(x)))\right|}
Applying properties of the absolute value and logarithms,
ln
|
h
(
x
)
|
=
ln
|
f
(
x
)
|
−
ln
|
g
(
x
)
|
{\displaystyle \ln |h(x)|=\ln |f(x)|-\ln |g(x)|}
Taking the logarithmic derivative of both sides,
h
′
(
x
)
h
(
x
)
=
f
′
(
x
)
f
(
x
)
−
g
′
(
x
)
g
(
x
)
{\displaystyle {\frac {h'(x)}{h(x)))={\frac {f'(x)}{f(x)))-{\frac {g'(x)}{g(x)))}
Solving for
h
′
(
x
)
{\displaystyle h'(x)}
and substituting back
f
(
x
)
g
(
x
)
{\displaystyle {\tfrac {f(x)}{g(x)))}
for
h
(
x
)
{\displaystyle h(x)}
gives:
h
′
(
x
)
=
h
(
x
)
[
f
′
(
x
)
f
(
x
)
−
g
′
(
x
)
g
(
x
)
]
=
f
(
x
)
g
(
x
)
[
f
′
(
x
)
f
(
x
)
−
g
′
(
x
)
g
(
x
)
]
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
=
f
′
(
x
)
g
(
x
)
−
f
(
x
)
g
′
(
x
)
g
(
x
)
2
.
{\displaystyle {\begin{aligned}h'(x)&=h(x)\left[{\frac {f'(x)}{f(x)))-{\frac {g'(x)}{g(x)))\right]\\&={\frac {f(x)}{g(x)))\left[{\frac {f'(x)}{f(x)))-{\frac {g'(x)}{g(x)))\right]\\&={\frac {f'(x)}{g(x)))-{\frac {f(x)g'(x)}{g(x)^{2))}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{g(x)^{2))}.\end{aligned))}
Taking the absolute value of the functions is necessary for the logarithmic differentiation of functions that may have negative values, as logarithms are only real-valued for positive arguments. This works because
d
d
x
(
ln
|
u
|
)
=
u
′
u
{\displaystyle {\tfrac {d}{dx))(\ln |u|)={\tfrac {u'}{u))}
, which justifies taking the absolute value of the functions for logarithmic differentiation.