This article is about a method in calculus. For other uses, see
Reciprocal .
In calculus , the reciprocal rule gives the derivative of the reciprocal of a function f in terms of the derivative of f . The reciprocal rule can be used to show that the power rule holds for negative exponents if it has already been established for positive exponents. Also, one can readily deduce the quotient rule from the reciprocal rule and the product rule .
The reciprocal rule states that if f is differentiable at a point x and f (x ) ≠ 0 then g(x ) = 1/f (x ) is also differentiable at x and
g
′
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x
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=
−
f
′
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x
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f
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2
.
{\displaystyle g'(x)=-{\frac {f'(x)}{f(x)^{2}}}.}
This proof relies on the premise that
f
{\displaystyle f}
x
,
{\displaystyle x,}
f
{\displaystyle f}
continuous there. Applying the definition of the derivative of
g
{\displaystyle g}
x
{\displaystyle x}
f
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x
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≠
0
{\displaystyle f(x)\neq 0}
g
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=
d
d
x
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1
f
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lim
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→
0
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1
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−
1
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lim
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→
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−
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h
⋅
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⋅
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=
lim
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→
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{\displaystyle {\begin{aligned}g'(x)={\frac {d}{dx}}\left({\frac {1}{f(x)}}\right)&=\lim _{h\to 0}\left({\frac {{\frac {1}{f(x+h)}}-{\frac {1}{f(x)}}}{h}}\right)\\&=\lim _{h\to 0}\left({\frac {f(x)-f(x+h)}{h\cdot f(x)\cdot f(x+h)}}\right)\\&=\lim _{h\to 0}\left(-({\frac {(f(x+h)-f(x)}{h}})\cdot ({\frac {1}{f(x)\cdot f(x+h)}})\right)\end{aligned}}}
(
−
lim
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→
0
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lim
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{\displaystyle \left(-\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\right)\cdot \left(\lim _{h\to 0}{\frac {1}{f(x)\cdot f(x+h)}}\right)}
f
{\displaystyle f}
x
{\displaystyle x}
−
f
′
(
x
)
,
{\displaystyle -f'(x),}
f
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x
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≠
0
{\displaystyle f(x)\neq 0}
f
{\displaystyle f}
x
{\displaystyle x}
equals
1
/
f
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x
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2
,
{\displaystyle 1/f(x)^{2},}
thus yielding
g
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−
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⋅
1
f
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2
=
−
f
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f
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2
{\displaystyle g'(x)=-f'(x)\cdot {\frac {1}{f(x)^{2}}}=-{\frac {f'(x)}{f(x)^{2}}}}
A weak reciprocal rule that follows algebraically from the product rule [ edit ] It may be argued that since
f
(
x
)
⋅
1
f
(
x
)
=
1
,
{\displaystyle f(x)\cdot {\frac {1}{f(x)}}=1,}
an application of the product rule says that
f
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1
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+
f
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1
f
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′
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=
0
,
{\displaystyle f'(x)\left({\frac {1}{f}}\right)(x)+f(x)\left({\frac {1}{f}}\right)'(x)=0,}
and this may be algebraically rearranged to say
(
1
f
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′
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x
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=
−
f
′
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x
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f
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2
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{\displaystyle \left({\frac {1}{f}}\right)'(x)={\frac {-f'(x)}{f(x)^{2}}}.}
However, this fails to prove that 1/f is differentiable at x ; it is valid only when differentiability of 1/f at x is already established. In that way, it is a weaker result than the reciprocal rule proved above. However, in the context of differential algebra , in which there is nothing that is not differentiable and in which derivatives are not defined by limits, it is in this way that the reciprocal rule and the more general quotient rule are established.
Application to generalization of the power rule [ edit ] Often the power rule, stating that
d
d
x
(
x
n
)
=
n
x
n
−
1
{\displaystyle {\tfrac {d}{dx}}(x^{n})=nx^{n-1}}
n is a nonnegative integer. This can be extended to negative integers n by letting
n
=
−
m
{\displaystyle n=-m}
m is a positive integer.
d
d
x
x
n
=
d
d
x
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1
x
m
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=
−
d
d
x
x
m
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m
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2
,
by the reciprocal rule
=
−
m
x
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−
1
x
2
m
,
by the power rule applied to the positive integer
m
,
=
−
m
x
−
m
−
1
=
n
x
n
−
1
,
by substituting back
n
=
−
m
.
{\displaystyle {\begin{aligned}{\frac {d}{dx}}x^{n}&={\frac {d}{dx}}\,\left({\frac {1}{x^{m}}}\right)\\&=-{\frac {{\frac {d}{dx}}x^{m}}{(x^{m})^{2}}},{\text{ by the reciprocal rule}}\\&=-{\frac {mx^{m-1}}{x^{2m}}},{\text{ by the power rule applied to the positive integer }}m,\\&=-mx^{-m-1}=nx^{n-1},{\text{ by substituting back }}n=-m.\end{aligned}}}
Application to a proof of the quotient rule [ edit ] The reciprocal rule is a special case of the quotient rule, which states that if f and g are differentiable at x and g (x ) ≠ 0 then
d
d
x
[
f
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g
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]
=
g
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f
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−
f
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g
′
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x
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[
g
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]
2
.
{\displaystyle {\frac {d}{dx}}\,\left[{\frac {f(x)}{g(x)}}\right]={\frac {g(x)f\,'(x)-f(x)g'(x)}{[g(x)]^{2}}}.}
The quotient rule can be proved by writing
f
(
x
)
g
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x
)
=
f
(
x
)
⋅
1
g
(
x
)
{\displaystyle {\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}}
and then first applying the product rule, and then applying the reciprocal rule to the second factor.
d
d
x
[
f
(
x
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g
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]
=
d
d
x
[
f
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⋅
1
g
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]
=
f
′
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⋅
1
g
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+
f
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⋅
d
d
x
[
1
g
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]
=
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1
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+
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[
−
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]
=
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g
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−
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g
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[
g
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2
=
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2
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{\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]&={\frac {d}{dx}}\left[f(x)\cdot {\frac {1}{g(x)}}\right]\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]\\&=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 {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}.\end{aligned}}}
Application to differentiation of trigonometric functions [ edit ] By using the reciprocal rule one can find the derivative of the secant and cosecant functions.
For the secant function:
d
d
x
sec
x
=
d
d
x
(
1
cos
x
)
=
−
d
d
x
cos
x
cos
2
x
=
sin
x
cos
2
x
=
1
cos
x
⋅
sin
x
cos
x
=
sec
x
tan
x
.
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\sec x&={\frac {d}{dx}}\,\left({\frac {1}{\cos x}}\right)={\frac {-{\frac {d}{dx}}\cos x}{\cos ^{2}x}}={\frac {\sin x}{\cos ^{2}x}}={\frac {1}{\cos x}}\cdot {\frac {\sin x}{\cos x}}=\sec x\tan x.\end{aligned}}}
The cosecant is treated similarly:
d
d
x
csc
x
=
d
d
x
(
1
sin
x
)
=
−
d
d
x
sin
x
sin
2
x
=
−
cos
x
sin
2
x
=
−
1
sin
x
⋅
cos
x
sin
x
=
−
csc
x
cot
x
.
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\csc x&={\frac {d}{dx}}\,\left({\frac {1}{\sin x}}\right)={\frac {-{\frac {d}{dx}}\sin x}{\sin ^{2}x}}=-{\frac {\cos x}{\sin ^{2}x}}=-{\frac {1}{\sin x}}\cdot {\frac {\cos x}{\sin x}}=-\csc x\cot x.\end{aligned}}}
![{\displaystyle {\frac {d}{dx}}\,\left[{\frac {f(x)}{g(x)}}\right]={\frac {g(x)f\,'(x)-f(x)g'(x)}{[g(x)]^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8796d779957e4cbcd36d6db3cbabb756de532d1)
![{\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]&={\frac {d}{dx}}\left[f(x)\cdot {\frac {1}{g(x)}}\right]\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]\\&=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 {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/beed5ab1507340a9c2a4b3be1568d392d5602b3f)
