The derivative dy/dx gives the gradient of a curve at any point. For a power of x, multiply by the power and reduce the power by one: if y = xⁿ, then dy/dx = nxⁿ⁻¹. So x⁴ becomes 4x³.
Differentiate a polynomial term by term. Numbers in front stay as multipliers, and a constant term disappears, because a horizontal line has gradient 0. For y = 4x³ − x + 7, dy/dx = 12x² − 1. Then substitute an x value to find the gradient at that point.
Worked example
f(x) = 5x³ + 5x + 3. Find f′(x) and the gradient at x = 1.
- f′(x) = 15x² + 5.
- f′(1) = 15 + 5 = 20.
Answer:f′(x) = 15x² + 5; gradient 20
Key words
- derivative
- the rate of change, or gradient function
- dy/dx
- notation for the derivative of y
- f′(x)
- the derivative of f(x)
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Common Core-referenced: HSF-IF.B.6