Reciprocal rule

This proof relies on the premise that ${\displaystyle f}$ is differentiable at ${\displaystyle x,}$ and on the theorem that ${\displaystyle f}$ is then also necessarily continuous there. Applying the definition of the derivative of ${\displaystyle g}$ at ${\displaystyle x}$ with ${\displaystyle f(x)\neq 0}$ gives

${\displaystyle {\begin{aligned}g'(x)={\frac {d}{dx}}\left({\frac {1}{f(x)}}\right)&=\lim _{h\to 0}\left({\frac {{\frac {1}{f(x+h)}}-{\frac {1}{f(x)}}}{h}}\right)\\&=\lim _{h\to 0}\left({\frac {f(x)-f(x+h)}{h\cdot f(x)f(x+h)}}\right)\\&=\lim _{h\to 0}\left(-{\frac {f(x+h)-f(x)}{h}}\cdot {\frac {1}{f(x)f(x+h)}}\right).\end{aligned}}}$

The limit of this product exists and is equal to the product of the existing limits of its factors:

${\displaystyle \left(\lim _{h\to 0}-{\frac {f(x+h)-f(x)}{h}}\right)\cdot \left(\lim _{h\to 0}{\frac {1}{f(x)\cdot f(x+h)}}\right).}$

Because of the differentiability of ${\displaystyle f}$ at ${\displaystyle x}$ the first limit equals ${\displaystyle -f'(x),}$ and because of ${\displaystyle f(x)\neq 0}$ and the continuity of ${\displaystyle f}$ at ${\displaystyle x}$ the second limit equals ${\displaystyle 1/f(x)^{2},}$ thus yielding

${\displaystyle g'(x)=-f'(x)\cdot {\frac {1}{f(x)^{2}}}=-{\frac {f'(x)}{f(x)^{2}}}.}$

It may be argued that since

${\displaystyle f(x)\cdot {\frac {1}{f(x)}}=1,}$

an application of the product rule says that

${\displaystyle f'(x)\left({\frac {1}{f}}\right)(x)+f(x)\left({\frac {1}{f}}\right)'(x)=0,}$

and this may be algebraically rearranged to say

${\displaystyle \left({\frac {1}{f}}\right)'(x)={\frac {-f'(x)}{f(x)^{2}}}.}$

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.

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

${\displaystyle {\begin{aligned}{\frac {d}{dx}}x^{n}&={\frac {d}{dx}}\,\left({\frac {1}{x^{m}}}\right)\\&=-{\frac {{\frac {d}{dx}}x^{m}}{(x^{m})^{2}}},{\text{ by the reciprocal rule}}\\&=-{\frac {mx^{m-1}}{x^{2m}}},{\text{ by the power rule applied to the positive integer }}m,\\&=-mx^{-m-1}=nx^{n-1},{\text{ by substituting back }}n=-m.\end{aligned}}}$

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

${\displaystyle {\frac {d}{dx}}\,\left[{\frac {f(x)}{g(x)}}\right]={\frac {g(x)f\,'(x)-f(x)g'(x)}{[g(x)]^{2}}}.}$

The quotient rule can be proved by writing

${\displaystyle {\frac {f(x)}{g(x)}}=f(x)\cdot {\frac {1}{g(x)}}}$

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

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]&={\frac {d}{dx}}\left[f(x)\cdot {\frac {1}{g(x)}}\right]\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot {\frac {d}{dx}}\left[{\frac {1}{g(x)}}\right]\\&=f'(x)\cdot {\frac {1}{g(x)}}+f(x)\cdot \left[{\frac {-g'(x)}{g(x)^{2}}}\right]\\&={\frac {f'(x)}{g(x)}}-{\frac {f(x)g'(x)}{[g(x)]^{2}}}\\&={\frac {f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}}.\end{aligned}}}$

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

For the secant function:

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\sec x&={\frac {d}{dx}}\,\left({\frac {1}{\cos x}}\right)={\frac {-{\frac {d}{dx}}\cos x}{\cos ^{2}x}}={\frac {\sin x}{\cos ^{2}x}}={\frac {1}{\cos x}}\cdot {\frac {\sin x}{\cos x}}=\sec x\tan x.\end{aligned}}}$

The cosecant is treated similarly:

${\displaystyle {\begin{aligned}{\frac {d}{dx}}\csc x&={\frac {d}{dx}}\,\left({\frac {1}{\sin x}}\right)={\frac {-{\frac {d}{dx}}\sin x}{\sin ^{2}x}}=-{\frac {\cos x}{\sin ^{2}x}}=-{\frac {1}{\sin x}}\cdot {\frac {\cos x}{\sin x}}=-\csc x\cot x.\end{aligned}}}$

• Product rule
• Quotient rule
• Chain rule
• Difference quotient