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Reciprocal Rule

Reciprocal Rule

A formula for the derivative of the reciprocal of a function.

 

Reciprocal Rule: (1/v)' = -v'/v². Examples with 1/(x³-4) and 1/sin(x) showing derivative calculations.

 

 

See also

Derivative rules

Key Formula

ddx[1f(x)]=−f′(x)[f(x)]2\frac{d}{dx}\left[\frac{1}{f(x)}\right] = -\frac{f'(x)}{\left[f(x)\right]^2}
Where:
  • f(x)f(x) = A differentiable function that is not equal to zero at the point of evaluation
  • f′(x)f'(x) = The derivative of f(x) with respect to x

Worked Example

Problem: Find the derivative of g(x) = 1/(x³) using the Reciprocal Rule.
Step 1: Identify f(x) inside the reciprocal. Here the function in the denominator is:
f(x)=x3f(x) = x^3
Step 2: Compute the derivative f'(x) using the power rule:
f′(x)=3x2f'(x) = 3x^2
Step 3: Apply the Reciprocal Rule formula by substituting f(x) and f'(x):
ddx[1x3]=−3x2(x3)2\frac{d}{dx}\left[\frac{1}{x^3}\right] = -\frac{3x^2}{(x^3)^2}
Step 4: Simplify the denominator and reduce the fraction:
=−3x2x6=−3x4= -\frac{3x^2}{x^6} = -\frac{3}{x^4}
Answer: g'(x) = -3/x⁴. You can verify this matches what you get by rewriting 1/x³ as x⁻³ and using the power rule: d/dx[x⁻³] = -3x⁻⁴ = -3/x⁴.

Another Example

This example applies the Reciprocal Rule to a trigonometric function rather than a polynomial, showing how the rule recovers the well-known derivative of csc(x).

Problem: Find the derivative of h(x) = 1/sin(x) using the Reciprocal Rule.
Step 1: Identify the function in the denominator:
f(x)=sin⁡(x)f(x) = \sin(x)
Step 2: Compute f'(x):
f′(x)=cos⁡(x)f'(x) = \cos(x)
Step 3: Apply the Reciprocal Rule:
ddx[1sin⁡(x)]=−cos⁡(x)sin⁡2(x)\frac{d}{dx}\left[\frac{1}{\sin(x)}\right] = -\frac{\cos(x)}{\sin^2(x)}
Step 4: Rewrite using trigonometric identities. Since 1/sin(x) = csc(x) and cos(x)/sin(x) = cot(x):
=−cos⁡(x)sin⁡(x)⋅1sin⁡(x)=−cot⁡(x)csc⁡(x)= -\frac{\cos(x)}{\sin(x)} \cdot \frac{1}{\sin(x)} = -\cot(x)\csc(x)
Answer: h'(x) = −cot(x)csc(x), which matches the standard derivative of csc(x).

Frequently Asked Questions

What is the difference between the Reciprocal Rule and the Quotient Rule?
The Reciprocal Rule is a special case of the Quotient Rule where the numerator is the constant 1. The Quotient Rule handles any ratio g(x)/f(x), giving [f(x)g'(x) − g(x)f'(x)] / [f(x)]². When g(x) = 1 (so g'(x) = 0), the Quotient Rule simplifies to −f'(x)/[f(x)]², which is exactly the Reciprocal Rule. Using the Reciprocal Rule when the numerator is 1 saves a step.
How do you derive the Reciprocal Rule?
You can derive it from the chain rule. Write 1/f(x) as [f(x)]⁻¹. By the chain rule, the derivative is −1·[f(x)]⁻² · f'(x), which equals −f'(x)/[f(x)]². Alternatively, set g(x) = 1 in the Quotient Rule and the same formula appears.
When should you use the Reciprocal Rule instead of the power rule?
If the denominator is a simple power of x, like 1/x⁵, you can rewrite it as x⁻⁵ and use the power rule directly. The Reciprocal Rule becomes more useful when the denominator is a composite or transcendental function, such as 1/sin(x) or 1/(x² + 1), where rewriting as a negative exponent would still require the chain rule anyway.

Reciprocal Rule vs. Quotient Rule

Reciprocal RuleQuotient Rule
DefinitionDerivative of 1/f(x)Derivative of g(x)/f(x) for any differentiable g and f
Formula−f'(x) / [f(x)]²[f(x)g'(x) − g(x)f'(x)] / [f(x)]²
When to useNumerator is the constant 1Numerator is any function of x
ComplexityOnly need f'(x) — one derivative to computeNeed both g'(x) and f'(x) — two derivatives to compute

Why It Matters

The Reciprocal Rule appears frequently in calculus courses whenever you need the derivative of expressions like 1/eˣ, 1/ln(x), or 1/cos(x). Recognizing when to use it simplifies your work compared to setting up the full Quotient Rule with a constant numerator. It also reinforces the chain rule, since the Reciprocal Rule is essentially the chain rule applied to [f(x)]⁻¹.

Common Mistakes

Mistake: Forgetting the negative sign in the formula, writing f'(x)/[f(x)]² instead of −f'(x)/[f(x)]².
Correction: The negative sign is essential. It comes from the exponent −1 in [f(x)]⁻¹. A quick sanity check: the derivative of 1/x should be −1/x², not 1/x². If your answer is positive where it should be negative, you likely dropped the sign.
Mistake: Squaring f'(x) in the denominator instead of squaring f(x).
Correction: The denominator is [f(x)]², not [f'(x)]². The square applies to the original function, not its derivative. For example, for 1/sin(x), the denominator is sin²(x), not cos²(x).

Related Terms