“KALKULUS PEUBAH BANYAK”

THE DERIVATIVE

Dosen Pengampu :
Eni Defitriani,S.pd,M.pd
Asisten dosen :
Martha Lestari, S.pd, M.pd

Disusun Oleh:
Najwa Nur Adha
(2300884202010)

PROGAM STUDI PENDIDIKAN MATEMATIKA

FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN
UNIVERSITAS BATANGHARI
2024
 THE DERIVATIVE
1.Tangent lines and Rates Of Change
A tangentis a straight line that touches a curveat a certain point.

DEFINISI Suppose that x 0is in the domain of the function f. The tangent line to thecurve y =
f(x) at the point. P(x 0 , f ( x0 ) ) is the line with equation

y−f ¿

Where
f ( x )−f (x 0)
mtan =lim
x → x0 x−x 0

Provided the limit exists. For simplicity, we will also call thisthe tangent line to y = f(x) at x 0

SLOPES AND RATES OF CHANGE

Velocity can be viewed as rate of change the rate of change of position with respect to
time. Rates of change occur in other applications as well. For example:
• A microbiologist might be interested in the rate at which the number of bacteria in a colony
changes with time
• An engineer might beinterested in the rate at which the length of a metal rod changes with
temperature.
• An economist might be interested in the rate at which production cost changes with the
quantity of a product that is manufactured.
• A medical researcher might be interested in the rate at which the radius of an artery changes
with the concentration of alcohol in the bloodstream.
Our next objective is to define precisely what is meant by the "rate of change of y
with respect to x” when y is a function of x. In the case where y is a linear function of x, say y
= mx + b the slope m is the natural measure of the rate of change of y with respect to x. As
illustrated in Figure 2.1.8, each 1-unit increase in x anywhere along the line produces an m-
unit change in y, so we see that y changesat a constant rate with respect to x along the line and
that m measures this rate of change.
Example
Suppose that a uniform rod of length 40 cm (0.4 m) is thermally insulated around the lateral
surface and that the exposed ends of the rod are held at constant temperatures of 25°C and
5°C, respectively (Figure 2.1.9a). It is shown in physics that under appropriate conditions the
graph of the temperature T versus the distance x from the left-hand end of the rod will be a
straight line. Parts (b) and (c) of Figure 2.1.9 show two such graphs: one in which x i
measured in centimeters and one in which it is measured in meters. The slopes in the two
cases are

The slope in (6) implies that the temperature decreasesat a rateof 0.5° C per centimeter of
distance from the left end of the rod, and the slope in (7) implies that the temperature
decreasesat a rate of 50° C per meter of distance from the left end of the rod. The two
statements are equivalent physically, even though the slopes differ.

2. The DerivativeFunction
f ( x +h )−f (x)
Definition The function f ' ( x )=¿ lim
h→ 0 h
Is called the derivative of f with respect to x. The domain of f’ consists of all x in the domain
of f for which the limit exist.
The term "derivative" is used because the function f' is derived from the function f by a
limiting process.
Example
Find the derivative with respect to x of f(x) = x² and use it to find the equation of the tangen
line to y = x²at x = 2

You can think of f' as a "slope-producing function" in the sense that the value of f'(x) at x = x 0
is the slope of the tangent line to thegraph of f at x = x 0This aspect of the derivative is
illustrated in Figure 2.2.2, which shows the graphs f(x) = x² and its derivative f’(x) = 2x
(obtained in Example 1). The figure illustrates that thevalues of f’(x) = 2x at dx = - 2, 0 and 2
correspond to the slopes of the tangent lines to the graph f(x) = x² at those values of x

3. Introduction to Techniques of Differentiation

DERIVATIVE OF A CONSTANT
 ' f ( x+ h )−f (x) c−c
f ( x )=lim =lim =lim 0=0
h →0 h h →0 h h→0

This, we have established the following result.

1. THEOREMA The derivative of a constant function is 0; thst is, if c is any real number,
d
then [ c ] =0
dx
d
Example: [ 1 ] =0 , d [−3 ] =0 , d [ π ]=0 , d [ − √2 ] =0
dx dx dx dx

2. THEOREM (The power rule) if n is a positive integer, then

d n
[ x ]=n x n−1
dx
d 4
Example: [ x ]=4 x 3 , d [ x 5 ]=5 x 4 , d [ t12 ]=12 t11
dx dx dx

3. THEOREM (Extended power rule) If r is any real number, then

d r
[ x ]=r x r−1
dx

Example
d π
[ x ]=π x π−1
dx
d 4 /5 4 ( 5 )−1 4 −1 /5
4

[ x ]= x = x
dx 5 5

4. THEOREM (Constant Multiple Rule) if f is differentiable at x and c is any real number,

then cf is also differentiable at x and
d d
[ cf ( x )] =c [ f ( x ) ]
dx dx

Example
d
[ 4 x8 ]=4 d [ x 8 ]=4 [ 8 x 7 ]=32 x 7
dx dx
d
[−x 12 ]=(−1 ) d [ x 12 ]=−12 x11
dx dx

5. THEOREM (Sum and Difference Rules) if f and g are differentiable at x, then so are
f + g and f – g and
 d d d
dx
[ f ( x ) + g(x ) ] = [ f (x) ] + [ g (x) ]
dx dx
d d d
dx
[ f ( x )−g ( x) ] = [ f (x) ] − [ g ( x) ]
dx dx

Example
d
1. [ 2 x6 + x −9 ]= d [ 2 x 6 ] + d [ x−9 ]=12 x 5 + (−9 ) x−10=12 x 5 −9 x−10
dx dx dx
2. Find dy/dx if y=3 x −2 x 5 +6 x+ 1
8

dy d
= [ 3 x −2 x +6 x +1 ]
8 5
dx dx
d
¿ [ 3 x 8 ]− d [ 2 x 5 ] + d [ 6 x ] + d [ 1 ]
dx dx dx dx
7 4
¿ 24 x −10 x +6

4. The Product and Quotient Rules

You might be tempted to conjecture that the derivative of a product of two functions is
the product of their derivatives. However, a simple example will show this to be false.
Consider the functions.
f (x)=x∧g (x)=x ²

The product of their derivatives is

f ' (x) g '(x )=(1)(2 x)=2 x

But their product is h ( x )=f ( x ) g ( x )=x 3so the derivative of the product is h '( x)=3 x ²

Thus, the derivative of the product is not equal to the product of the derivatives. The correct
relationship, which is credited to Leibniz, is given by the following theorem.

THEOREM (The Product Rule) If f and g are differentiable at x, then so is th eproducts

f . g , and
d d d
dx
[ f ( x ) g (x) ]= f ( x ) [ g ( x ) ] + g ( x) [ f (x ) ]
dx dx

Example

Find dy/dx if y=(4 x ²−1)(7 x ³ + x)

Solution. There are two methods that can be used to find dy/dx. We can either use the product
rule or we can multiply out the factors in y and then differentiate. We will give both methods.

DERIVATIVE OF A QUOTIENT
Just as the derivative of a productis not generally the product of the derivatives, so the
derivative of a quotientis not generally the quotient of the derivatives. The correct
relationship is given by the following theorem.

THEOREM (The Patient Rale) If f and g are both differentiable at x and if g(x )≠ 0 then
f /g is differentiable at x and
d d
g(x) [ f ( x ) ] −f (x ) [ g( x ) ]
[ ]
d f (x )
dx g( x)
=
dx
[ g(x )]
2
dx

SUMMARY OF DIFFERENTIATION RULES

The following table summarizes the differentiation rules that we have encountered thus far
5. Derivatives of Trigonometric Functions
We will assume in thissection that the variable x in the trigonometric functions
sin x ,cos x , tan x , cot x , sec x ,∧csc x is measured in radians.

Example
Find dy/dx if y=x sin x
Solution. Using Formula (3) and the product rule we obtain
dy d
= [ x sin x ]
dx dx
 d
¿x [ sin x ] + sin x d [ x ]
dx dx
¿ x cos +sin x

The derivatives of the remaining trigonometric functions are

d d
[ tan x ] =sec 2 x [ sec x ]=sec x tan x
dx dx

d d
[ cot x ] =−csc 2 x [ csc x ] =−csc x cot x
dx dx

6. The Chain Rule

THEOREM (The Chain Rule) if g is differentiable at x and f is differentiable at g(x), then the
composition f ° g is differentiable at x. Moreover, if

y=f ( g ( x )) ∧u=g (x)

Then y=f (u) and

dy dy du
= .
dx du dx

Example: Find dy/dx if y=cos ⁡( x3 )

Solution. Let u=x3 and express y as y=cos u. Applying formula (1) yields
dy dy du
= .
dx du dx
d d
¿ [ cos u ] . [ x 3 ]
du dx
2
¿ (−s∈u ) .(3 x )

¿ (−sin ( x ) ) . ( 3 x )=−3 x sin(x )

3 2 2 3

AN ALTERNATIVE VERSION OF THE CHAIN RULE

d
dx
[ f (g ( x ) )]=( f o g )' =f ' ( g ( x ) ) g' (x )
A convenient way to remember this formula is to call f the "outside function" and the "inside
function" in the composition f ( g ( x ) )and then express (2) in words as:

The derivative of f(g(x)) is the derivative of the outside function evaluated at the inside
function times the derivative of the inside function.

Example

Find h ’ (x)if h(x )=cos (x ³)

Solution. We can think of h as a composition f (g(x )) in which g(x )=x ³ is the inside
function and f (x)=cos x is the outsite function

GENERALIZED DERIVATIVE FORMULAS

There is a useful third variation of the chain rule that strikes a middle ground between
formulas (1) and (2). If we let u=g (x) in (2), then we can rewrite that formula as
d du
[ f (u) ] =f ' (u)
dx dx
This result, called the generaslized derivative formula for f, provides a way of using the
derivative of f(x) to produce the derivative of f(u), where u is a function of x. Table 2.6.1
gives some examples of this formula