Высшая математика: пределы, непрерывность, дифференцирование. Левичев А.В - 16 стр.

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Solution
Find the s -increment:
∆s =
1
2
g(t +∆t)
2
−
1
2
gt
2
= gt∆t +
1
2
g(∆t)
2
.
Arrange for the difference quotient:
∆s
∆t
= gt +
1
2
g∆t. Find the
limit of the difference quotient as ∆t goes to zero:
v = lim
∆t→0
∆s
∆t
= gt.
Answer:v = gt; the value of this function (at t =2)is19.6m/sec.
4.2. Definition of the Derivative
Definition 16
The derivative of a function y = f(x) at x = a, denoted by
f
(a) is
f
(a) = lim
h→0
f(a + h) − f(a)
h
or f
(a) = lim
∆x→0
∆y
∆x
if this limit exists.
One can use x for a,and∆x for h to get f
(x), derivative at x.
In that case,
f
(x) = lim
∆x→0
f(x +∆x) − f(x)
∆x
= lim
∆x→0
∆y
∆x
.
Clearly, f
(x) is, in itself, a function of x. It should always be
clear from the context whether we look for a number f
(a) (in the
case x = a is a chosen number) or for a function f
(x) (when we allow
a to vary). Also, f
(a) is read: ”f prime of a“ (Newton’s notation for
the derivative).
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