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CIVIL ENGINEERING FACULTY  MUHAMMA DIYAH SURAKARTA UNIVERSIT Y 2010 THE DERIVATIVE By ALFIA MAGFIRONA

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LIST OF SLIDE

A. DEFINITION

B. THE DERIVATIVE FUNCTIONS

C. PARAMETRIC FUNCTION

2

D. IMPLICIT FUNCTION

E. APLICATION OF DERVATIVE

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A. DEFINITION

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What is the derivative ?

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A. DEFINITION

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In calculus  (a branch of mathematics) thederivative  is a measure of how a function 

changes as its input changes. Loosely speaking, aderivative can be thought of as how much onequantity is changing in response to changes insome other quantity; for example, the derivative

of the position of a moving object with respect totime is the object's instantaneous velocity. 

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A. DEFINITION

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The derivative of a function at a chosen input valuedescribes the best linear approximation  of the functionnear that input value. For a real-valued function  of asingle real variable, the derivative at a point equals theslope of the tangent line  to the  graph of the function atthat point. In higher dimensions, the derivative of afunction at a point is a linear transformation called thelinearization.[1] A closely related notion is the differential

of a function. The process of finding a derivative is calleddifferentiation.. 

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A. DEFINITION

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It can be shown by the graphic bellow :

The graph of a function, drawn in black, and atangent line to that function, drawn in red. The slopeof the tangent line is equal to the derivative of thefunction at the marked point.

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A. DEFINITION

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 Various Symbols for the Derivative

( )  or '( ) ordy df x  f xdx dx

0

Definition: lim x

dy y

dx x

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A. DEFINITION

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Higher Order Derivatives

Let y  = f ( x ). We have:

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Some Basic Derivatives

A. DEFINITION

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A. DEFINITION

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A. DEFINITION

Chain Rule 

The last formula

is known as the Chain Rule formula. It may be

rewritten as

 Another similar formula is given by

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B. THE DERIVATVE FUNCTIONS

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Derivative of Trigonometry

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B. THE DERIVATVE FUNCTIONS

1. = 4 + 2,  

… 

= 44 − 22 

2.  = 2 3 + 4,

… 

= 2 23 + 4. 32 

= 62. 23 + 4 

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B. THE DERIVATVE FUNCTIONS

erivative of the Hyperbolic functions and theirInverses

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B. THE DERIVATVE FUNCTIONS

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B. THE DERIVATVE FUNCTIONS

Recall the definitions of the trigonometricfunctions

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Derivative of the Exponential and Logarithmic functions

B. THE DERIVATVE FUNCTIONS

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EXAMPLE

B. THE DERIVATVE FUNCTIONS

  = 24 + 3,

……?

Solution :

1) 

Remember the Formula :()

=  −1

 

(24 + 3)

= 2. ln4 + 3.(4 + 3

 

2)  After that, differentiate( 4+3

, by using the formula :

(ln)

=1

 

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B. THE DERIVATVE FUNCTIONS(ln(4+3))

=

1

(4+3). 3=

3

(4+3) 

Finally, from the explanation above we can solve the ques tion

by :

24+3

= 2.ln4+3.(4+3

 

= 2. ln4+3.1

(4+3)4+ 3

 

= 2. ln4+3.3

(4+3) 

=6 ln(4+3)(4+3)

 

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Derivative of yclometry

B. THE DERIVATVE FUNCTIONS

In mathematics, the inverse trigonometric functions or cyclometric

functions are the inverse functions of the trigonometric functions.

The principal inverses are listed in the following table.

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B. THE DERIVATVE FUNCTION

NameUsual

notationDefinition

Domain of  x  

for real result

Range ofusual

principal

value

(radians)

Range ofusual

principal

value

(degrees)

arcsine  y  = arcsin  x    x  = sin y   −1 ≤  x  ≤ 1  −π/2 ≤ y  ≤ π/2  −90° ≤ y  ≤ 90°

arccosine  y  = arccos  x    x  = cos y   −1 ≤  x  ≤ 1  0 ≤ y  ≤ π  0° ≤ y  ≤ 180°

arctangent  y  = arctan  x    x  = tan y  all real

numbers−π/2 < y  < π/2  −90° < y  < 90°

arccotangent  y  = arccot  x    x  = cot y  all real

numbers0 < y  < π  0° < y  < 180°

arcsecant  y  = arcsec  x    x  = sec y    x  ≤ −1 or 1 ≤  x  0 ≤ y  < π/2 or

π/2 < y  ≤ π 

0° ≤ y  < 90° or

90° < y  ≤ 180°

arccosecant  y  = arccsc  x    x  = csc y    x  ≤ −1 or 1 ≤  x  −π/2 ≤ y  < 0

or 0 < y  ≤ π/2 

-90° ≤ y  < 0°

or 0° < y  ≤ 90°

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EXAMPLE

B. THE DERIVATVE FUNCTIONS

=   (2 − 5),……? 

Answer :

=−1

 1− (2 −5)2.(2 − 5)

 

=−1

 1−(2

−10+25)

. 2 

=−2

 −2+10−24 

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C. PARAMETRIC DERIVATIVE

The common form of parametric function :

x = f(t)

y = g(t) (t as a parameter)

To differentiate the function in the form of parametric, take :

x = f(t), so

=  ′

() 

y = g(t), so 

= ′() 

It can be exlplained :()

()=

=

.

 

  FORMULA x = f(t)

y = g(t)

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if :

EXAMPLE

x = 2t2 + t

y = t5 + 1

find : a). b).

C. PARAMETRIC DERIVATIVE

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Solution

). = 22 + ,   = 4 + 1 

= 5 + 1,   = 54 

=

= 5

4

4 + 1 

). = 54,  

22 = 203 

= 4

+ 1,

2

2= 4 

2 2

=

2 2

2 2

3 =

4+1203 .54.44+13 =

203(3+1)

4+13  

C. PARAMETRIC DERIVATIVE

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Implicit erivativeIn many examples, especially the ones derived from differential

equations, the variables involved are not linked to each other in an

explicit way. Most of the time, they are linked through an implicitformula, like F ( x ,y ) =0. Once  x   is fixed, we may find y   through

numerical computations. (By some fancy theorems, we may formally

show that y   may indeed be seen as a function of  x   over a certain

interval). The question becomes what is the derivative , at least at a

certain a point? The method of implicit differentiation answers thisconcern. Let us illustrate this through the following example. 

D. IMPLICIT DERIVATIVE

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D. IMPLICIT DERIVATIVE

If there is implicit function : 3 + + 3 = 0,   … ? 

Solution :

3 + + 3 = 0 

32 + + (.

+ 32 .

) = 0 

+ 32.

= −(32 + ) 

=−32 +

+ 32 

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E. APLICATION DERIVATIVE

A patient is being fed inntravenously from a conical-shapedbottle of maximum radius 4 inchies and 10 inchies. If the

rate of flow is 28 cubic inchies per hour find the rate at

which the fluid level is dropping in the bottle when the level

is at 8 inchies; when the level is at 2 inchies.

Solution

We again follow the procedure outline in Remark 2.

1. The units indicated in the problem are inchies and hours.

From the sketch in Fig. 1 and the cross section in Fig 2 wechoose the variables r=radius of fluid in cone and h= height

of fluid in cone.

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E. APLICATION DERIVATIVE

4

10 10

h

r

4

Fig. 1 Fig 2 

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E. APLICATION DERIVATIVE2. In term of these variables, we need to find

, the

rate at which the height of the fluid is changing. The

rate of flow is given to be :

=  −28    

Where V=volume of fluid in the cone.

3. The volume of fluid is given by the formula for the

volume of a cone.

=2

3

 

One of the variables can be eliminated by the

relationship between corresponding parts of similar

triangle in Fig.2

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E. APLICATION DERIVATIVEThat is,

4=

10 

Substituting = 0,4, we have

=

(0,4)2

3 =

0,163

3  

So that

=

0,16(32)

3= 0,162 

4. From the chain rule :

=

 

E APLICATION DERIVATIVE

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E. APLICATION DERIVATIVEAnd so,

=

=

−28

0.162 

In particular, when = 8, 

=8 =

−28

0,16(8)2 =−0,87(  ) 

When h = 2

=2 =−28

0,16(2)2= −13,93(  ) 

Notice that the fluid level is dropping much more rapidly as the cone empties

even though the flow rate is a constant 28 cubic inchies per hour. You can

observe this phenomenon by watching as it empties.

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BIBLIOGRAPHY

http://en.wikipedia.org/wiki/Derivative

http://www.sosmath.com/.../derivative.html 

http://www.sosmath.com/.../der05.html  

http://en.wikipedia.org/wiki/Inverse_trigonometric 

L. Mett, Corren & C. Smith, James. 1991. Calculus With

Aplication . York Graphich Services, Inc. USA.

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