The definition of the derivative is

lim f(x + Δx) - f(x)

Δx→ 0 Δx

We can therefore get the derivative of f(x) = 2x² + 5x using the above as follows:

lim [ 2(x + Δx)² + 5(x + Δx) – (2x² + 5x)]

Δx → 0 Δx

lim [2(x² + 2xΔx + Δx²) + 5x + 5Δx – 2x² - 5x]

Δx → 0 Δx

lim 2x² + 5x + 4xΔx + 2Δx² + 5Δx - 2x² – 5x

Δx→ 0 Δx

lim 4xΔx + 2Δx² + 5Δx

Δx→ 0 Δx

We can now factor out a Δx in the numerator to get

lim Δx(4x + 2 Δx + 5)

Δx→ 0 Δx

Δx in the numerator and denominator cancel out to get

lim 4x + 2 Δx + 5

Δx→ 0

Substituting 0 for Δx gives us 4x + 5, which is the derivative.

There is a much easier way to get the derivative than using the formal definition.

Multiply the coefficient by the exponent in the first term. That result becomes the new coefficient and subtract one from the exponent to get the new exponent

For 2x² that is 2(2) = 4 (new coefficient), exponent goes from 2 to (2-1) = 1. The first term of the derivative is 4x

For 5x the is 5(1) = 5 (new coefficient), exponent goes from 1 to (1-1) = 0. x⁰ = 1, so there is no x in the second term.

The derivative is 4x + 5