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Derivative of a Power Series — Formula & Examples

Derivative of a Power Series

The derivative of a function defined by a power series can be found by differentiating the series term-by-term.

 

Power series f(x)=Σc_n(x−a)^n and its derivative f'(x)=Σnc_n(x−a)^(n−1), both converging on interval (c,d).

 

 

See also

Integral of a power series, convergent series, divergent series, differentiable

Key Formula

If f(x)=∑n=0∞an(x−c)n,then f′(x)=∑n=1∞n an(x−c)n−1\text{If } f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n, \quad \text{then } f'(x) = \sum_{n=1}^{\infty} n\, a_n (x - c)^{n-1}
Where:
  • f(x)f(x) = The function defined by the power series
  • ana_n = The coefficient of the nth term
  • cc = The center of the power series
  • xx = The variable
  • nn = The index of summation, starting from 1 (since the n=0 constant term vanishes)

Worked Example

Problem: Find the derivative of the power series f(x) = ∑ (from n=0 to ∞) xⁿ/n!, which represents eˣ.
Step 1: Write out the power series explicitly.
f(x)=∑n=0∞xnn!=1+x+x22!+x33!+⋯f(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots
Step 2: Differentiate each term using the power rule. The constant term 1 drops away, and each xⁿ becomes nxⁿ⁻¹.
f′(x)=∑n=1∞n xn−1n!f'(x) = \sum_{n=1}^{\infty} \frac{n\, x^{n-1}}{n!}
Step 3: Simplify by canceling n with n! to get (n−1)!.
f′(x)=∑n=1∞xn−1(n−1)!f'(x) = \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}
Step 4: Re-index by letting m = n − 1, so the sum starts at m = 0.
f′(x)=∑m=0∞xmm!=exf'(x) = \sum_{m=0}^{\infty} \frac{x^m}{m!} = e^x
Answer: f'(x) = eˣ, confirming the well-known result that the derivative of eˣ is itself.

Another Example

This example uses a series with non-trivial coefficients (n+1) rather than simple factorials, and shows how term-by-term differentiation of a known series can generate the series for a more complex closed-form function.

Problem: Find the derivative of f(x) = ∑ (from n=0 to ∞) (n+1)x^n centered at c = 0.
Step 1: Recognize what function this series represents. Recall that 1/(1−x)² = ∑ (n+1)xⁿ for |x| < 1. So f(x) = 1/(1−x)².
f(x)=∑n=0∞(n+1)xn=1+2x+3x2+4x3+⋯f(x) = \sum_{n=0}^{\infty}(n+1)x^n = 1 + 2x + 3x^2 + 4x^3 + \cdots
Step 2: Differentiate term-by-term. Apply the rule: bring down the exponent n and reduce the power by 1.
f′(x)=∑n=1∞n(n+1) xn−1f'(x) = \sum_{n=1}^{\infty} n(n+1)\,x^{n-1}
Step 3: Write out a few terms to see the pattern.
f′(x)=2+6x+12x2+20x3+⋯f'(x) = 2 + 6x + 12x^2 + 20x^3 + \cdots
Step 4: This equals 2/(1−x)³, which you can verify by differentiating f(x) = (1−x)⁻² directly using the chain rule.
f′(x)=2(1−x)3f'(x) = \frac{2}{(1-x)^3}
Answer: f'(x) = ∑ (from n=1 to ∞) n(n+1)xⁿ⁻¹ = 2/(1−x)³, valid for |x| < 1.

Frequently Asked Questions

Does the radius of convergence change when you differentiate a power series?
No. A power series and its term-by-term derivative always have the same radius of convergence R. However, convergence behavior at the endpoints x = c ± R can change. A series that converges at an endpoint may have a derivative series that diverges there, or vice versa.
Can you differentiate a power series more than once?
Yes. A power series is infinitely differentiable inside its interval of convergence. You can differentiate term-by-term repeatedly to find f''(x), f'''(x), and so on, each time producing a new power series with the same radius of convergence.
Why does the summation index start at n = 1 after differentiating?
The n = 0 term of the original series is just the constant a₀, and the derivative of a constant is zero. So that term disappears, and the first nonzero term in the derivative series corresponds to n = 1.

Derivative of a Power Series vs. Integral of a Power Series

Derivative of a Power SeriesIntegral of a Power Series
OperationDifferentiate each term: aₙxⁿ → n·aₙxⁿ⁻¹Integrate each term: aₙxⁿ → aₙxⁿ⁺¹/(n+1)
Index shiftSummation starts at n = 1 (constant term drops out)Summation keeps n = 0; a constant of integration C is added
Radius of convergenceSame as original seriesSame as original series
Endpoint behaviorMay lose convergence at endpointsMay gain convergence at endpoints
Effect on coefficientsMultiplies aₙ by n (amplifies higher-order terms)Divides aₙ by (n+1) (dampens higher-order terms)

Why It Matters

Term-by-term differentiation of power series is essential in solving differential equations, particularly in the Frobenius method and when finding series solutions near ordinary or singular points. It also lets you derive new series from known ones — for example, differentiating the geometric series 1/(1−x) yields 1/(1−x)², giving you its power series without computing coefficients from scratch. In physics and engineering, this technique is used constantly when working with Taylor and Maclaurin series to approximate derivatives of complicated functions.

Common Mistakes

Mistake: Forgetting to change the starting index from n = 0 to n = 1 after differentiating.
Correction: The n = 0 term is a constant and its derivative is zero, so the differentiated series must begin at n = 1. Leaving the index at n = 0 can introduce a spurious term or cause an undefined expression like 0 · x⁻¹.
Mistake: Assuming the interval of convergence (including endpoints) is identical for the original and differentiated series.
Correction: The radius of convergence stays the same, but endpoint behavior can differ. Always recheck convergence at x = c + R and x = c − R separately after differentiating.

Related Terms