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Last updated on July 23rd, 2025

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Derivative of Factorial

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Understanding the derivative of a factorial function can be quite complex, as the factorial is traditionally defined only for integers. However, by extending the concept to the Gamma function, which generalizes the factorial to non-integer values, we can explore its derivative. Derivatives play a crucial role in various fields, including mathematics and engineering. We will now discuss the derivative of factorial in detail.

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What is the Derivative of Factorial?

The derivative of the factorial function is not straightforward since the factorial is only defined for non-negative integers.

 

To find a derivative, we use the Gamma function, Γ(n), which extends the factorial function to complex and real numbers.

 

The Gamma function satisfies Γ(n) = (n-1)! for natural numbers.

 

Thus, the derivative of the factorial can be explored through the derivative of the Gamma function.

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Derivative of Factorial Formula

The derivative of the Gamma function, which extends the factorial, is given by: d/dx [Γ(x)] = Γ(x)ψ(x) where ψ(x) is the digamma function, the logarithmic derivative of the Gamma function.

 

This formula applies to all values where the Gamma function is defined.

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Proofs of the Derivative of Factorial

To understand the derivative of the factorial through the Gamma function, we explore it using differentiation techniques.

 

The derivative of Γ(x) = ∫₀^∞ t^(x-1)e^(-t)dt can be differentiated under the integral sign, using the Leibniz rule.

 

The derivative involves the digamma function, which is the derivative of the logarithm of the Gamma function.

 

Using Differentiation Under the Integral Sign The Gamma function Γ(x) = ∫₀^∞ t^(x-1)e^(-t)dt can be differentiated with respect to x: d/dx [Γ(x)] = ∫₀^∞ ∂/∂x [t^(x-1)e^(-t)] dt = ∫₀^∞ t^(x-1) ln(t) e^(-t) dt

 

This result involves the digamma function, ψ(x), such that: d/dx [Γ(x)] = Γ(x)ψ(x)

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Higher-Order Derivatives of Factorial

Higher-order derivatives of the Gamma function can be derived similarly to those of any other function.

 

Considering the first derivative involves the digamma function, higher-order derivatives involve polygamma functions, denoted as ψ^(n)(x).

 

These derivatives become progressively complex, but they provide deeper insights into the properties of the Gamma function and thus the factorial.

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Special Cases:

For integer values, the Gamma function's derivative simplifies considerably.

 

However, at non-positive integers, the Gamma function is undefined, reflecting the factorial function's nature.

 

For instance, at x = 0, the Gamma function has a pole, and its derivative is undefined.

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Common Mistakes and How to Avoid Them in Derivatives of Factorial

Mistakes often occur when applying derivatives to factorials due to misunderstanding the transition from factorials to the Gamma function.

 

Here are a few common mistakes and ways to resolve them:

Mistake 1

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Misapplying the Gamma Function

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Students may incorrectly apply the Gamma function, forgetting that it generalizes the factorial to non-integers.

 

To avoid this, ensure a clear understanding of how the Gamma function extends the factorial concept.

Mistake 2

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Forgetting the Domain of the Gamma Function

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The Gamma function is undefined at non-positive integers.

 

Students should remember that attempting to differentiate at these points will result in undefined expressions.

Mistake 3

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Ignoring the Digamma Function

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When differentiating the Gamma function, failing to incorporate the digamma function leads to incorrect results. Always include the digamma function, ψ(x), in your differentiation to ensure accuracy.

Mistake 4

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Confusing Gamma and Factorial

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Confusing the notations and definitions of the Gamma function and factorial can lead to errors.

 

Remember, Γ(n) = (n-1)! for integers n, and the Gamma function extends beyond integers.

Mistake 5

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Overlooking Higher-Order Derivatives

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Ignoring the complexity of higher-order derivatives can result in incomplete analyses.

 

Recognize that higher-order derivatives involve polygamma functions, which are derivatives of the digamma function.

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Examples Using the Derivative of Factorial

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Problem 1

Calculate the derivative of Γ(x) for x = 3.

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To find the derivative of Γ(x) at x = 3, we use the formula: d/dx [Γ(x)] = Γ(x)ψ(x) Given that Γ(3) = 2! = 2, we need to find ψ(3). ψ(3) can be calculated using known values or tables, and it equals approximately 0.922784.

 

Thus, d/dx [Γ(x)] at x = 3 is: Γ(3)ψ(3) = 2 * 0.922784 ≈ 1.845568.

Explanation

We find the derivative of the Gamma function at x = 3 by using the digamma function's value. The Gamma function at x = 3 gives 2!, and multiplying this by ψ(3) provides the derivative value.

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Problem 2

A company models its production rate with a function based on factorials extended by the Gamma function. Given the function f(x) = Γ(x) for production, find the rate of change in production when x = 5.

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To find the rate of change in production, we differentiate f(x) = Γ(x): d/dx [Γ(x)] = Γ(x)ψ(x) For x = 5, Γ(5) = 4! = 24. ψ(5) can be found from tables or known values, approximately 1.506118.

 

Thus, the rate of change in production when x = 5 is: Γ(5)ψ(5) = 24 * 1.506118 ≈ 36.146832.

Explanation

The rate of change in production is found by differentiating the Gamma function and using the digamma function value at x = 5.

 

We multiply Γ(5) with ψ(5) to obtain the result.

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Problem 3

Derive the second derivative of the function f(x) = Γ(x).

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The first derivative is: d/dx [Γ(x)] = Γ(x)ψ(x)

 

To find the second derivative, differentiate again: d²/dx² [Γ(x)] = d/dx [Γ(x)ψ(x)]

 

Using the product rule: = Γ(x)ψ(x)² + Γ(x)ψ'(x)

 

The second derivative involves the trigamma function ψ'(x).

Explanation

We derive the second derivative by differentiating the first derivative, using the product rule to account for the digamma and trigamma functions, resulting in a more complex expression.

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Problem 4

Prove: d/dx [Γ²(x)] = 2Γ(x)Γ(x)ψ(x).

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Using the chain rule: Let y = Γ²(x) y = [Γ(x)]²

 

Differentiate using the chain rule: dy/dx = 2Γ(x) d/dx [Γ(x)] Since d/dx [Γ(x)] = Γ(x)ψ(x), dy/dx = 2Γ(x)Γ(x)ψ(x)

 

Thus, d/dx [Γ²(x)] = 2Γ(x)Γ(x)ψ(x) is proved.

Explanation

In this proof, we apply the chain rule to differentiate Γ²(x), using the derivative of Γ(x) to find the desired result.

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Problem 5

Solve: d/dx [Γ(x)/x].

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Using the quotient rule: d/dx [Γ(x)/x] = (x d/dx [Γ(x)] - Γ(x) d/dx [x]) / x² Substitute d/dx [Γ(x)] = Γ(x)ψ(x) and d/dx [x] = 1: = (xΓ(x)ψ(x) - Γ(x)) / x² = (Γ(x)(xψ(x) - 1)) / x² Therefore, d/dx [Γ(x)/x] = (Γ(x)(xψ(x) - 1)) / x².

Explanation

The differentiation uses the quotient rule, substituting the derivative of Γ(x) and simplifying to achieve the final expression.

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FAQs on the Derivative of Factorial

1.Find the derivative of the Gamma function.

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2.Can the derivative of the factorial be used in real life?

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3.Is the derivative of the Gamma function defined at negative integers?

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4.What rule is used to differentiate Γ(x)/x?

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5.Are the derivatives of the Gamma function and factorial the same?

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Important Glossaries for the Derivative of Factorial

  • Gamma Function: A function that extends the factorial to non-integer values, denoted as Γ(x).

 

  • Digamma Function: The derivative of the logarithm of the Gamma function, represented as ψ(x).

 

  • Quotient Rule: A method used to differentiate ratios of functions.

 

  • Higher-Order Derivatives: Derivatives taken multiple times, involving more complex functions like polygamma.

 

  • Polygamma Function: Derivatives of the digamma function, denoted ψ^(n)(x), representing higher-order derivatives of the Gamma function.
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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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