View source for Reciprocal rule

{{Confuse|Reciprocal formula (circuit networks)}}
In [[Calculus|calculus]], the '''reciprocal rule''' gives the derivative of the [[Multiplicative inverse|reciprocal]] of a function ''f'' in terms of the derivative of&nbsp;''f''. The reciprocal rule can be used to show that the [[Power rule|power rule]] holds for negative exponents if it has already been established for positive exponents. Also, one can readily deduce the [[Quotient rule|quotient rule]] from the reciprocal rule and the [[Product rule|product rule]].

The reciprocal rule states that if ''f'' is [[Differentiable function|differentiable]] at a point ''x'' and ''f''(''x'')&nbsp;&ne;&nbsp;0 then g(''x'') = 1/''f''(''x'') is also differentiable at ''x'' and

<math display=block> g'(x) = -\frac{f'(x)}{f(x)^2}. </math>

== Proof ==

This proof relies on the premise that <math>f</math> is differentiable at <math>x,</math> and on the theorem that <math>f</math> is then also necessarily [[Continuous function|continuous]] there. Applying the definition of the derivative of <math>g</math> at <math>x</math> with <math>f(x) \ne 0</math> gives
<math display=block>\begin{align}
 g'(x) = \frac d {dx} \left(\frac {1} {f(x)}\right) & = \lim_{h\to0} \left (\frac{\frac {1}{f(x+h)} - \frac {1}{f(x)}} {h} \right )\\
& = \lim_{h\to0} \left( \frac{f(x)-f(x+h)}{h \cdot f(x)f(x+h)} \right)\\
& = \lim_{h\to0} \left( - \frac{f(x+h)-f(x)}{h} \cdot \frac 1 {f(x)f(x+h)} \right).\end{align}</math>
The limit of this product exists and is equal to the product of the existing limits of its factors:
<math display=block>\left( \lim_{h\to0} -\frac{f(x+h)-f(x)}{h} \right)\cdot\left( \lim_{h\to0} \frac {1}{f(x)\cdot f(x+h)}\right).</math>
Because of the differentiability of <math>f</math> at <math>x</math> the first limit equals <math>-f'(x),</math> and because of <math>f(x)\ne 0</math> and the continuity of <math>f</math> at <math>x</math> the second limit {{nowrap|equals <math>1/f(x)^2,</math>}} thus yielding
<math display=block>g'(x) = -f'(x) \cdot \frac {1}{f(x)^2} = -\frac{f'(x)}{f(x)^2}.</math>

== A weak reciprocal rule that follows algebraically from the product rule ==

It may be argued that since

<math display=block> f(x)\cdot \frac 1 {f(x)} = 1, </math>

an application of the product rule says that

<math display=block> f'(x) \left( \frac 1 f\right)(x) + f(x) \left( \frac 1 f\right)'(x) = 0, </math>

and this may be algebraically rearranged to say

<math display=block> \left( \frac 1 f\right)'(x) = \frac{-f'(x)}{f(x)^2}. </math>

However, this fails to prove that 1/''f'' is differentiable at&nbsp;''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|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 ==

Often the power rule, stating that <math>\tfrac{d}{dx}(x^n) = nx^{n-1}</math>, is proved by methods that are valid only when ''n'' is a nonnegative integer. This can be extended to negative integers ''n'' by letting <math>n = -m</math>, where ''m'' is a positive integer.

<math display=block>
\begin{align}
\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{align}
</math>

== Application to a proof of the quotient rule ==

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'')&nbsp;&ne;&nbsp;0 then

<math display=block> \frac d {dx}\, \left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f\,'(x) - f(x)g'(x)}{[g(x)]^2}. </math>

The quotient rule can be proved by writing

<math display=block> \frac{f(x)}{g(x)} = f(x) \cdot \frac 1 {g(x)} </math>

and then first applying the product rule, and then applying the reciprocal rule to the second factor.

<math display=block>
\begin{align}
\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{align}
</math>

== Application to differentiation of trigonometric functions ==

By using the reciprocal rule one can find the derivative of the secant and cosecant functions.

For the secant function:

<math display=block>
\begin{align}
\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{align}
</math>

The cosecant is treated similarly:

<math display=block>
\begin{align}
\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{align}
</math>

==See also==

* {{annotated link|Chain rule}}
* {{annotated link|Difference quotient}}
* {{annotated link|Differentiation of integrals}}
* {{annotated link|Differentiation rules}}
* {{annotated link|General Leibniz rule}}
* {{annotated link|Integration by parts}}
* {{annotated link|Inverse functions and differentiation}}
* {{annotated link|Linearity of differentiation}}
* {{annotated link|Product rule}}
* {{annotated link|Quotient rule}}
* {{annotated link|Vector calculus identities}}

==References==

{{reflist}}

{{Calculus topics}}

[[Category:Articles containing proofs]]
[[Category:Differentiation rules]]
[[Category:Theorems in analysis]]
[[Category:Theorems in calculus]]

{{Sourceattribution|Reciprocal rule}}