Last updated: 2026-05-26

Derivative of a polynomial at a point

The derivative of a polynomial P(x) at a point x₀ is the instantaneous rate of change of P at that point, written P′(x₀). It equals the slope of the line tangent to the curve y = P(x) at (x₀, P(x₀)). Because every term of a polynomial is a power of x with a constant coefficient, the derivative can be computed exactly by the power rule, term by term.

This calculator takes the polynomial's coefficients from highest to lowest degree (for example, 1, -3, 2, 5 for x³ − 3x² + 2x + 5) and a value of x, and returns P(x), P′(x), the coefficients of the derivative polynomial, and the equation of the tangent line, with a graph of the curve and its tangent.

Scope: Polynomial inputs only. Symbolic differentiation of arbitrary expressions (sin, exp, ln, etc.) is not supported.

The Power Rule

For any term aₙxⁿ, the derivative is:

$\frac{d}{dx}[a_n x^n] = n \cdot a_n x^{n-1}$

Applied term by term to a polynomial:

$P(x) = a_0 x^n + a_1 x^{n-1} + \cdots + a_{n-1}x + a_n$

$P'(x) = n\,a_0\, x^{n-1} + (n-1)\,a_1\, x^{n-2} + \cdots + a_{n-1}$

The constant term vanishes because $\tfrac{d}{dx}[c] = 0$, and the degree of the derivative is one less than the degree of the original polynomial.

Worked Example

Take P(x) = x³ − 3x² + 2x + 5 (coefficients: 1, -3, 2, 5) evaluated at x = 2.

Step 1 — Evaluate P(2) using Horner's method:

$P(2) = ((1 \cdot 2 - 3) \cdot 2 + 2) \cdot 2 + 5 = 5$

Step 2 — Differentiate using the power rule:

$P'(x) = 3x^2 - 6x + 2$

Step 3 — Evaluate P′(2):

$P'(2) = 3(4) - 6(2) + 2 = 12 - 12 + 2 = 2$

Step 4 — Write the tangent line:

$y = P'(2)\,(x - 2) + P(2) = 2(x - 2) + 5 = 2x + 1$

The tangent line touches the curve at (2, 5) with slope 2.

Geometric Meaning of the Derivative

The derivative P′(x₀) is the instantaneous rate of change of the polynomial at x₀ — equivalently, the slope of the curve at that single point. As the interval [x₀, x₀ + h] shrinks to zero, the secant slope converges to P′(x₀):

$P'(x_0) = \lim_{h \to 0} \frac{P(x_0 + h) - P(x_0)}{h}$

The tangent line is the best linear approximation to the curve near x₀. For small departures Δx, the curve and the tangent line diverge by an amount proportional to (Δx)², which is why the tangent is an excellent local model of the polynomial.

Entering Coefficients

Important: include zero coefficients for missing powers. For x⁴ − 1, the x³, x², and x¹ terms are all zero, so the input is 1, 0, 0, 0, -1.

Derivative Coefficients Output

The calculator also returns the coefficients of P′(x) in the same high-to-low format. For P(x) = x³ − 3x² + 2x + 5:

$P'(x) = 3x^2 - 6x + 2 \quad \Rightarrow \quad \text{coefficients: } 3, -6, 2$

These coefficients can be pasted back into the input to differentiate again (finding P″(x)), or used in the Quadratic Equation Solver to find where P′(x) = 0 (the critical points of P). The Average Rate of Change Calculator computes the secant slope between two points — the discrete analogue of the derivative; as the two points approach each other, the secant slope converges to P′(x₀).