Mechanics 1: Linear Kinematics and Calculus - 3D Math Primer for Graphics and Game Development - page 499

Differentiating a simple function by using the definition Equation (11.3)
is an important rite of passage, and we are proud to help you cross this
threshold. The typical procedure is this:

1. Substitute x(t) and x(t+ ∆t) into the definition. (In our case, x(t) =
−t 2 + 6t − 1).

2. Perform algebraic manipulations until it is legal to substitute ∆t = 0.
(Often this boils down to getting ∆t out of the denominator.)

3. Substitute ∆t = 0, which evaluates the expression “at the limit,”
removing the limit notation.

4. Simplify the result.

Applying this procedure to our case yields

v(t) = dx

x(t + ∆t) − x(t)
∆t dt = lim

∆t→0

[−(t + ∆t) 2 + 6(t + ∆t) − 1] − (−t 2 + 6t − 1)
∆t

= lim
∆t→0

(−t 2 − 2t(∆t) − (∆t) 2 + 6t + 6(∆t) − 1) + (t 2 − 6t + 1)
∆t

= lim
∆t→0

−2t(∆t) − (∆t) 2 + 6(∆t)
∆t

= lim
∆t→0

∆t(−2t − ∆t + 6)
∆t

= lim
∆t→0

= lim
∆t→0

−2t − ∆t + 6. (11.4)

Now we are at step 3. Taking the limit in Equation (11.4) is now easy; we
simply substitute ∆t = 0. This substitution was not legal earlier because
there was a ∆t in the denominator:
v(t) = dx

dt = lim

−2t − ∆t + 6

∆t→0

= −2t − (0) + 6
= −2t + 6. (11.5)

Finally! Equation (11.5) is the velocity function we've been looking for.
It allows us to plug in any value of t and compute the instantaneous velocity
of the hare at that time. Putting in t = 2.5, we arrive at the answer to our
question:

v(t) = −2t + 6,
v(2.5) = −2(2.5) + 6 = 1.