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]
f
(
x
)
=
g
(
x
)
/
h
(
x
)
,
{\displaystyle f(x)=g(x)/h(x),}
g and h are differentiable and
h
(
x
)
≠
0.
{\displaystyle h(x)\neq 0.}
f (x )
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
h
(
x
)
2
.
{\displaystyle f'(x)={\frac {g'(x)h(x)-g(x)h'(x)}{h(x)^{2}}}.}
It is provable in many ways by using other differential rules.
Examples
A basic example:
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
=
e
x
(
x
−
2
)
x
3
.
{\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}}}
The quotient rule can be used to find the derivative of
f
(
x
)
=
tan
x
=
sin
x
cos
x
{\displaystyle f(x)=\tan x={\tfrac {\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
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}}{\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}}}
Proofs
Proof from derivative definition and limit properties
Let
f
(
x
)
=
g
(
x
)
h
(
x
)
.
{\displaystyle f(x)={\frac {g(x)}{h(x)}}.}
f
′
(
x
)
=
lim
k
→
0
f
(
x
+
k
)
−
f
(
x
)
k
=
lim
k
→
0
g
(
x
+
k
)
h
(
x
+
k
)
−
g
(
x
)
h
(
x
)
k
=
lim
k
→
0
g
(
x
+
k
)
h
(
x
)
−
g
(
x
)
h
(
x
+
k
)
k
⋅
h
(
x
)
h
(
x
+
k
)
=
lim
k
→
0
g
(
x
+
k
)
h
(
x
)
−
g
(
x
)
h
(
x
+
k
)
k
⋅
lim
k
→
0
1
h
(
x
)
h
(
x
+
k
)
=
(
lim
k
→
0
g
(
x
+
k
)
h
(
x
)
−
g
(
x
)
h
(
x
)
+
g
(
x
)
h
(
x
)
−
g
(
x
)
h
(
x
+
k
)
k
)
⋅
1
h
(
x
)
2
=
(
lim
k
→
0
g
(
x
+
k
)
h
(
x
)
−
g
(
x
)
h
(
x
)
k
−
lim
k
→
0
g
(
x
)
h
(
x
+
k
)
−
g
(
x
)
h
(
x
)
k
)
⋅
1
h
(
x
)
2
=
(
h
(
x
)
lim
k
→
0
g
(
x
+
k
)
−
g
(
x
)
k
−
g
(
x
)
lim
k
→
0
h
(
x
+
k
)
−
h
(
x
)
k
)
⋅
1
h
(
x
)
2
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
h
(
x
)
2
.
{\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}}}
Proof using implicit differentiation
Let
f
(
x
)
=
g
(
x
)
h
(
x
)
,
{\displaystyle f(x)={\frac {g(x)}{h(x)}},}
g
(
x
)
=
f
(
x
)
h
(
x
)
.
{\displaystyle g(x)=f(x)h(x).}
product rule then gives
g
′
(
x
)
=
f
′
(
x
)
h
(
x
)
+
f
(
x
)
h
′
(
x
)
.
{\displaystyle g'(x)=f'(x)h(x)+f(x)h'(x).}
f
′
(
x
)
{\displaystyle f'(x)}
f
(
x
)
{\displaystyle f(x)}
f
′
(
x
)
=
g
′
(
x
)
−
f
(
x
)
h
′
(
x
)
h
(
x
)
=
g
′
(
x
)
−
g
(
x
)
h
(
x
)
⋅
h
′
(
x
)
h
(
x
)
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
h
(
x
)
2
.
{\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}}}
Proof using the chain rule
Let
f
(
x
)
=
g
(
x
)
h
(
x
)
=
g
(
x
)
h
(
x
)
−
1
.
{\displaystyle f(x)={\frac {g(x)}{h(x)}}=g(x)h(x)^{-1}.}
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
−
1
+
g
(
x
)
⋅
d
d
x
(
h
(
x
)
−
1
)
.
{\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot {\frac {d}{dx}}(h(x)^{-1}).}
power rule along with the chain rule :
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
−
1
+
g
(
x
)
⋅
(
−
1
)
h
(
x
)
−
2
h
′
(
x
)
.
{\displaystyle f'(x)=g'(x)h(x)^{-1}+g(x)\cdot (-1)h(x)^{-2}h'(x).}
f
′
(
x
)
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
h
(
x
)
2
=
g
′
(
x
)
h
(
x
)
−
g
(
x
)
h
′
(
x
)
h
(
x
)
2
.
{\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}}}
Implicit differentiation can be used to compute the n th derivative of a quotient (partially in terms of its first n − 1
f
h
=
g
{\displaystyle fh=g}
f
″
h
+
2
f
′
h
′
+
f
h
″
=
g
″
{\displaystyle f''h+2f'h'+fh''=g''}
f
″
{\displaystyle f''}
f
″
=
(
g
h
)
″
=
g
″
−
2
f
′
h
′
−
f
h
″
h
.
{\displaystyle f''=\left({\frac {g}{h}}\right)''={\frac {g''-2f'h'-fh''}{h}}.}
See also
References
