functions - administrative mathematics

solution:

The chain rule is used because it has a function within a function to a higher power .


Let the function f ( x ) = (4x3 +3 x2 - 2x4) 3


Applying the chain rule

where :


Y = (4x3 +3 x2 - 2x4) 3


Dy / du = 3 (4x3 +3 x2 - 2x4) 2


U = (4x3 +3 x2 - 2x4)


Du / dx = ( 12x2 +6 x - 8x3 )


First derivative of the function is calculated within the parentheses , and multiplied by the derivative of the outside .


F ' ( x ) = 3 (4x3 - 2x4 +3 x2 ) 2 ( +6 x 12x2 - 8x3 )


F '( x ) = ( 36x2 + 18x - 24x3 ) (4x3 +3 x2 - 2x4) 2


Conclusion : The derivative of the function is: F '( x ) = 36x2 + 18x - 24x3 ) (4x3 +3 x2 - 2x4) 2






Formula



Solution: the quotient rule is used u = 3x4 - x2 du / dx = 12x3 - 2x (applying derivative of a sum


 v = x3 +6 x2 dv / dx = 3x2 + 6x (applying derived from the sum )


Substituting in the formula, the derivative of g ( x ) function with respect to x : g '( x ) , is :

f '( x ) = ( x3 +6 x2 ) ( 12x2 - 2x ) - ( 3x4 - x2 ) ( 12x + 3x2 )

       -------------------------------------------------- -

                            ( x3 +6 x2 ) 2


f '( x ) = 12x6 - 2x4 + 72x5 - 12x3 - (9 x6 + x5 36 - 3x4 -12 x3 )

       -------------------------------------------------- ----------------------

                                   ( x3 +6 x2 ) 2


f '( x ) = 12x6 - 2x4 + 72x5 - 12x3 - 9 x6 - 36 x5 + 3x4 - 12x3

       -------------------------------------------------- ----------------------

                                   ( x3 +6 x2 ) 2


f '( x ) = 3x6 + x5 + 1 x4 36

       -------------------------------

              ( x3 +6 x2 ) 2

     

Conclusion : the derivative of the function is: 3x6 + x4 36 x5 + 1

-------------------------------

    ( x3 +6 x2 ) 2





solution:


u = 5x du / dx = 5


v = x3 - 62x +1 dv / dx = applying the formula dau / dx = lna au du / dx where a = 6 u = 2x -x3 +1 du / dx = 2 - 3x2


dv / dx = 62x -x3 +1 (2 - 3x2) 6 Ln


Applying the formula


F '( x ) = 5x [ 62x -x3 +1 (2 - 3x2) Ln 6] 5 + ( 62x -x3 +1 )


Simplifying the equation :


F '( x ) = 5 ( 62x -x3 +1 ) [ x (2 - 3x2) Ln 6 +1 ]


Conclusion : The results of this derivative is = F '( x ) = 5 ( 62x -x3 +1 ) [ x (2 - 3x2) Ln 6 +1 ]





Solution: Use the formula

       Where u = 2x4 + 2x2 -1

U ' = 8x3 + 4x



G ( x ) = 8x3 + 4x

        ----------------

           2x4 + 2x2 -1

     

 

                                                    1

X '= Dx [ln 2x4 + 2x2 -1] -------------------- Dx = [ 2x4 + 2x2 -1] =

( 2x4 + 2x2 -1 )


X '= 8x3 + 4x

      -------------

    ( 2x4 + 2x2 -1 )


Conclusion : It is obtained that the derivative of the function is then g ( x ) = 8x3 + 4x

--------------------

( 2x4 + 2x2 -1 )



May .


solution:


Let the function ( 3x +1 ) 3. The quotient rule is used which is equal to the derivative of the numerator by the denominator least the derivative of

                         ---------- Denominator by the numerator divided by the denominator squared

                           2x +2


Formula is used:






where

 u = ( 3x +1 ) 3 du / dx = 9 ( 3x +1 ) 2

Applying the chain rule y = ( 3x +1 ) 3 dy / du = 3 ( 3x +1 ) 2

U = ( 3x +1 ) du / dx = 3

Dy / dx = 3 ( 3x +1 ) 2 (3 ) = 9 ( 3x +1 ) 2

v = 2x +2 dv / dx = 2


Replacing data in the formula


F ' ( x ) = ( 2x + 2 ) [ ( 9) ( 3x +1 ) 2] - ( 2) ( 3x +1 ) 3

           --------------------------------------------

( 2x +2) 2

simplifying operations


F ' ( x ) = (18x +18) ( 3x +1 ) 2 - [ ( 3x +1 ) 3 (2 ) ]

           -------------------------------------------

( 2x +2) 2


Conclusion : The derivative of the function is f '(x ) = ) = (18x +18) ( 3x +1 ) 2 - (2 ) ( 3x +1 ) 3 / ( 2x +2) 2




    By Logarithmic difference.

solution:


Logarithmic laws are used



                        ( 3x +1 ) 3

Ln f '( x ) = ln y = ----------------- = Ln = ln ( 3x +1 ) 2 - Ln ( 2x +2)

2x +2


3ln Ln y = ( 3x +1) - ln ( 2x +2)


The function of the denominator is raised to a power , it is multiplied by the same power and function of the denominator remains the same as it is not raised to a power .

1

Applying the logarithmic formulated dLnu = dy = ------ du

or

d Lny

------- = 3LN (3x +1) - Ln2x +2

   dx

















By applying the corresponding derivatives are obtained


Dy 2dx 9DX

---- = ----------------------------

2x +2 and 3x +1


Both components contain dx, the factor we have:


Dy September 2

---- = dx -----------------------

2x +2 and 3x +1





Solving for y and dx for the derivative of the formula we have:


Dy September 2

---- = ----------------------- And

3x dx +1 2x +2


Is replaced and the original function is represented , obtaining the expected result.


  Dy September 2 ( 3x +1 ) 3

F ' ( x ) = ------ = ------------------------------------- -----

+2 dx 2x 2x 3x +2 +1



Conclusion : The derivative of the function is:


  Dy September 2 ( 3x +1 ) 3

F ' ( x ) = ------ = ------------------------------------- -----

+2 dx 2x 2x 3x +1 +2






Exercise 2 . Real income from marginal revenue


Solve the following problem.


    Revenue in store construction materials for the sale of sand are given by the following function:



In thousands of pesos weekly , determine the actual income from the sale of the bag of sand 101.


answer:


to determine the actual income from the sale of the 101 bag of sand in the store building materials , the concept of marginal revenue is used

Formula is used:

i ( x +1) - i ( x )

We want to know the product so we have 101 WHAT:

x +1 = 101 , x -1 = 101-1 = 100 x = 100

replace the function i ( x ) = 200x +30 in the formula of marginal revenue is :

i '= ( 300 ( x +1 ) +30 ) - ( 300 ( x ) +30 ) , the value of x = 100 is replaced

i '= ( 300 ( 100 +1 ) +30 ) - ( 300 ( 100 ) +30 )

i '= ( 300 ( one hundred and one ) +30 ) - ( 30030 )

i '= ( +30 30300 ) - ( 30030 )

i '= 30330-30030

i '= 30330-30030 i ' = 300



Conclusion : therefore real incomes by selling 101 bag of sand , would be $ 300,000 per week


_ DLn (2x +2) =


U = 2x +2

du = 2dx

  2dx

---------

2x +2


Since it was replaced in the formula proceeds to perform operations by multiplying the first variable variables in the second parentheses after the variable in the first parenthesis is multiplied by the second parentheses and the same is done in parentheses after sign out less , is hoisted factor and go adding or subtracting variables that have the same exponent for simplicity.


U = ( 3x +1 )

du = 3

3 (3) dx

---------

3x +1


  9DX

---------

3x +1


U = 2x4 + 2x2 -1

First searching the derivative of u, which is obtained by multiplying the variable 2x4 exponent to which this high at this CaSO4 and to apply the formula the formula un = nun -1 for the exponent to which this elevated variable resulting 8x3 and the same procedure is performed with the second variable being 4x and -1 in the derived variable is 0 . Resulting from the derivative of u.

u '= 8x3 + 4x


Data are substituted into the formula and the result is obtained taking the derivative of the denominator u ( u ' ) , and the denominator u.




d Lny 1

------- = -------- = D3ln ( 3x +1 ) =

dx and


The values ​​of u, du / dx, v , dv / dx, for the latter variable is going to take into account only what is inside the parentheses searching the values ​​of a, u, du / dx Wanted , using the formula:


Dau du

------ = ------- Au Lna

   dx dx


Data are substituted into the formula and simplifies