Derivative of the Reciprocal of a Function
Let \( f(x) \) be a differentiable function that is never equal to zero. The derivative of its reciprocal is obtained by taking the negative of the derivative of the function and dividing it by the square of the function itself:
\[ \left(\frac{1}{f(x)}\right)'= -\frac{f'(x)}{[f(x)]^2} \]
This formula is valid only at points where \( f(x)\neq 0 \), since the reciprocal of a function is undefined whenever the denominator is zero.
The rule depends on two key quantities:
- the rate of change of the function, represented by \( f'(x) \)
- the value of the function itself, which appears squared in the denominator
The negative sign shows that the reciprocal changes in the opposite direction to the original function. As \( f(x) \) increases, its reciprocal tends to decrease. Likewise, as \( f(x) \) decreases, its reciprocal tends to increase.
Example
Find the derivative of the function
\[ y=\frac{1}{\sin x} \]
The function in the denominator is
\[ f(x)=\sin x \]
Its derivative is
\[ f'(x)=\cos x \]
Applying the reciprocal rule:
\[ y' = -\frac{\cos x}{\sin^2 x} \]
Example 2
Consider the function
\[ y=\frac{5}{x^3-2} \]
Begin by factoring out the constant coefficient:
\[ y=5\cdot\frac{1}{x^3-2} \]
The denominator function is
\[ f(x)=x^3-2 \]
and its derivative is
\[ f'(x)=3x^2 \]
Substituting into the formula gives
\[ y' = 5\left( -\frac{3x^2}{(x^3-2)^2} \right) \]
Therefore,
\[ y' = -\frac{15x^2}{(x^3-2)^2} \]
Proof Using the Definition of the Derivative
Consider the function
\[ y=\frac{1}{f(x)} \]
Using the definition of the derivative:
\[ y'= \lim_{h\to0} \frac{\frac{1}{f(x+h)}-\frac{1}{f(x)}}{h} \]
Rearranging the expression:
\[ y'= \lim_{h\to0} \left(\frac{1}{f(x+h)}-\frac{1}{f(x)}\right)\frac{1}{h} \]
Combine the fractions over a common denominator:
\[ y'= \lim_{h\to0} \frac{f(x)-f(x+h)}{f(x)f(x+h)} \cdot \frac{1}{h} \]
Rewrite the expression as
\[ y'= \lim_{h\to0} \frac{f(x)-f(x+h)}{h}\cdot\frac{1}{f(x)f(x+h)} \]
Factor out the negative sign:
\[ y'= \lim_{h\to0} -1\cdot\left(\frac{f(x+h)-f(x)}{h}\right)\cdot\frac{1}{f(x)f(x+h)} \]
Since \( -1 \) is a constant, it can be moved outside the limit:
\[ y'= -\lim_{h\to0}\left[\left(\frac{f(x+h)-f(x)}{h}\right)\cdot\frac{1}{f(x)f(x+h)}\right] \]
Applying the product law for limits:
\[ y'= -\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\cdot\lim_{h\to0}\frac{1}{f(x)f(x+h)} \]
The first limit is, by definition, the derivative of \( f(x) \):
\[ y'= -\underbrace{\lim_{h\to0}\frac{f(x+h)-f(x)}{h}}_{f'(x)}\cdot\lim_{h\to0}\frac{1}{f(x)f(x+h)} \]
Substituting \( f'(x) \) gives
\[ y'= -f'(x)\cdot\lim_{h\to0}\frac{1}{f(x)f(x+h)} \]
Because every differentiable function is continuous,
\[ \lim_{h\to0}f(x+h)=f(x) \]
Therefore,
\[ y'= -f'(x)\cdot\lim_{h\to0}\frac{1}{f(x)\cdot f(x)} \]
\[ y'= -f'(x)\cdot\lim_{h\to0}\frac{1}{f^2(x)} \]
The second limit is now independent of \( h \), so
\[ y'= -f'(x)\cdot\frac{1}{f^2(x)} \]
Finally,
\[ y'=-\frac{f'(x)}{f^2(x)} \]
which proves the formula.
Proof Using the Power Rule
The reciprocal rule can also be obtained directly from the power rule.
Notice that the reciprocal of a function can be written as a negative power:
\[ \frac{1}{f(x)}=[f(x)]^{-1} \]
Therefore,
\[ \left(\frac{1}{f(x)}\right)' = \left([f(x)]^{-1}\right)' \]
Using the power rule for a composite function,
\[ \frac{d}{dx}[f(x)]^n=n[f(x)]^{n-1}f'(x) \]
and setting \( n=-1 \), we obtain
\[ \frac{d}{dx}[f(x)]^{-1}=(-1)[f(x)]^{-2}f'(x) \]
\[ \frac{d}{dx}[f(x)]^{-1}=-[f(x)]^{-2}f'(x) \]
\[ \frac{d}{dx}[f(x)]^{-1}=-\frac{f'(x)}{[f(x)]^2} \]
This is exactly the formula for the derivative of the reciprocal of a function.
Note. This derivation shows that the reciprocal rule is not an independent differentiation formula. Instead, it follows directly from the general power rule applied to the exponent \( n=-1 \). For this reason, many calculus textbooks present it as a special case of differentiating a composite function raised to a power.
Proof Using the Quotient Rule
The same result can also be obtained from the quotient rule.
Start with
\[ \left(\frac{1}{f(x)}\right)' \]
and define
\[ u(x)=1 \]
Then
\[ \left(\frac{u(x)}{f(x)}\right)' \]
Applying the quotient rule gives
\[ \left(\frac{u(x)}{f(x)}\right)'=\frac{u'(x)f(x)-u(x)f'(x)}{[f(x)]^2} \]
Since \( u(x)=1 \) is a constant,
\[ u'(x)=0 \]
Substituting into the formula:
\[ \left(\frac{u(x)}{f(x)}\right)'=\frac{0\cdot f(x)-u(x)\cdot f'(x)}{[f(x)]^2} \]
\[ \left(\frac{u(x)}{f(x)}\right)'=\frac{-u(x)\cdot f'(x)}{[f(x)]^2} \]
Since \( u(x)=1 \),
\[ \left(\frac{1}{f(x)}\right)'=\frac{-1\cdot f'(x)}{[f(x)]^2} \]
Simplifying the expression yields
\[ \left(\frac{1}{f(x)}\right)'=-\frac{f'(x)}{[f(x)]^2} \]
This completes the derivation.
