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]
h
(
x
)
=
f
(
x
)
g
(
x
)
{\displaystyle h(x)={\frac {f(x)}{g(x)))}
f and g are differentiable and
g
(
x
)
≠
0.
{\displaystyle g(x)\neq 0.}
h (x )
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 .
Example 1: Basic example [ edit ] Given
h
(
x
)
=
e
x
x
2
{\displaystyle h(x)={\frac {e^{x)){x^{2))))
f
(
x
)
=
e
x
,
g
(
x
)
=
x
2
{\displaystyle f(x)=e^{x},g(x)=x^{2))
d
d
x
(
e
x
x
2
)
=
(
d
d
x
e
x
)
(
x
2
)
−
(
e
x
)
(
d
d
x
x
2
)
(
x
2
)
2
=
(
e
x
)
(
x
2
)
−
(
e
x
)
(
2
x
)
x
4
=
x
2
e
x
−
2
x
e
x
x
4
=
x
e
x
−
2
e
x
x
3
=
e
x
(
x
−
2
)
x
3
.
{\displaystyle {\begin{aligned}{\frac {d}{dx))\left({\frac {e^{x)){x^{2))}\right)&={\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 {x^{2}e^{x}-2xe^{x)){x^{4))}\\&={\frac {xe^{x}-2e^{x)){x^{3))}\\&={\frac {e^{x}(x-2)}{x^{3))}.\end{aligned))}
Example 2: Derivative of tangent function [ edit ] The quotient rule can be used to find the derivative of
tan
x
=
sin
x
cos
x
{\displaystyle \tan x={\frac {\sin x}{\cos x))}
d
d
x
tan
x
=
d
d
x
(
sin
x
cos
x
)
=
(
d
d
x
sin
x
)
(
cos
x
)
−
(
sin
x
)
(
d
d
x
cos
x
)
cos
2
x
=
(
cos
x
)
(
cos
x
)
−
(
sin
x
)
(
−
sin
x
)
cos
2
x
=
cos
2
x
+
sin
2
x
cos
2
x
=
1
cos
2
x
=
sec
2
x
.
{\displaystyle {\begin{aligned}{\frac {d}{dx))\tan x&={\frac {d}{dx))\left({\frac {\sin x}{\cos x))\right)\\&={\frac {\left({\frac {d}{dx))\sin x\right)(\cos x)-(\sin x)\left({\frac {d}{dx))\cos x\right)}{\cos ^{2}x))\\&={\frac {(\cos x)(\cos x)-(\sin x)(-\sin x)}{\cos ^{2}x))\\&={\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x))\\&={\frac {1}{\cos ^{2}x))=\sec ^{2}x.\end{aligned))}
The reciprocal rule is a special case of the quotient rule in which the numerator
f
(
x
)
=
1
{\displaystyle f(x)=1}
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.
Proof from derivative definition and limit properties [ edit ] Let
h
(
x
)
=
f
(
x
)
g
(
x
)
.
{\displaystyle h(x)={\frac {f(x)}{g(x))).}
f
(
x
)
g
(
x
)
{\displaystyle f(x)g(x)}
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))}
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))))
g
(
x
)
{\displaystyle g(x)}
lim
k
→
0
g
(
x
+
k
)
=
g
(
x
)
{\displaystyle \lim _{k\to 0}g(x+k)=g(x)}
Proof using implicit differentiation [ edit ] Let
h
(
x
)
=
f
(
x
)
g
(
x
)
,
{\displaystyle h(x)={\frac {f(x)}{g(x))),}
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)}
h
(
x
)
{\displaystyle h(x)}
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 [ edit ] 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 [ edit ] Let
h
(
x
)
=
f
(
x
)
g
(
x
)
.
{\displaystyle h(x)={\frac {f(x)}{g(x))).}
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)}
f
(
x
)
g
(
x
)
{\displaystyle {\tfrac {f(x)}{g(x)))}
h
(
x
)
{\displaystyle h(x)}
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))}
Higher order derivatives [ edit ] Implicit differentiation can be used to compute the n th derivative of a quotient (partially in terms of its first n − 1
f
=
g
h
{\displaystyle f=gh}
f
″
=
g
″
h
+
2
g
′
h
′
+
g
h
″
{\displaystyle f''=g''h+2g'h'+gh''}
h
″
{\displaystyle h''}
h
″
=
(
f
g
)
″
=
f
″
−
g
″
h
−
2
g
′
h
′
g
.
{\displaystyle h''=\left({\frac {f}{g))\right)''={\frac {f''-g''h-2g'h'}{g)).}
![{\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))}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4e609f5e7b44f69edc0c1030bc9ae9d0cb2a104)
![{\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))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39ff7267ba5a87cdd17ba664f2983de34c26b434)
![{\displaystyle h'(x)=f'(x)\cdot {\frac {1}{g(x)))+f(x)\cdot {\frac {d}{dx))\left[{\frac {1}{g(x)))\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4eb1ec4c4cd3ce8891ece4a7f4db407fd158470c)
![{\displaystyle {\frac {d}{dx))\left[{\frac {1}{g(x)))\right]=-{\frac {1}{g(x)^{2))}\cdot g'(x)={\frac {-g'(x)}{g(x)^{2))}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3578de83a411d6ecc2753ffb336bbefa818b2508)
![{\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))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98814d0c465bffd44efe44e56bb1a84dd7fb1529)
![{\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))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a30778dbc5e1cd7b1f21fec351752fc348cb88af)
