The definition of the derivative is

lim

__f(x +____Δ____x) - f(x)__
Δx→
0 Δx

We
can therefore get the derivative of f(x) = 2x

^{2}+ 5x using the above as follows:
lim

__[ 2(x +____Δ____x)__^{2}__+ 5(x +____Δ____x) – (2x__^{2}__+ 5x)]__
Δx
→ 0 Δx

lim

__[2(x__^{2}__+ 2x____Δ____x +____Δ____x__^{2}__) + 5x + 5____Δ____x – 2x__^{2}__- 5x]__
Δx
→ 0 Δx

lim

__2x__^{2 }__+ 5x + 4x____Δ____x + 2____Δ____x__^{2}__+ 5____Δ____x - 2x__^{2}__– 5x__
Δx→
0 Δx

lim

__4x____Δ____x + 2____Δ____x__^{2}__+ 5____Δ____x__
Δx→
0 Δx

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

lim

__Δ____x(4x + 2____Δ____x + 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
2x

^{2}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. x

^{0}= 1, so there is no x in the second term.
The
derivative is 4x + 5

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