Variable Selection

The variable selection represents that the process of assigning the variable to some values and functions. The differentiation process sometimes involve big functions on that time the reference variable is used to assign and then it process into the simple calculations. In this way some confusions exist for the students for which variable we can ssign and for which functions. In this article we are going to discuss about the variable selection process in detail with the applications of the differentiation and integration process.

Examples for the variable selection in integration

• Review on the variable selection with the integrable function  `int tanh^-1 2theta ` `d theta`    

Solution:

Variable Selection:

Here the confusion exist for the variable selection.

Only one function exist  `tanh^-1 2theta`        

The another function is the constant term   1 d`theta`                   

Another one confusion

For which term we can take it as u and dv

If we choose u then it should be easily differentiable.

If we choose dv then it should be easily integrable.

u = `tanh^-1 2theta` and `(du)/(d theta)` = `theta` ;   dv = 1 d`theta`

du =` 1/(1-(2theta)^2) 2 d theta = 2/(1-4theta^2) d theta ` and  v = `theta` .

Therefore,

`int tanh^-1 2theta ` = `theta`   `tanh^-1 2theta` – `int theta 2/(1-4theta^2)` d`theta`   = `theta`   `tanh^-1 2theta` – `2int theta 1/(1-4theta^2)` d`theta` .

Let     u = `1-4theta^2`  

 `(du)/(d theta)` = -8`theta` ,

`(-1/8)` du = `theta` d`theta` .

 `int tanh^-1 2theta ` = `theta`   `tanh^-1 2theta` – `2int theta 1/(1-4theta^2)` d`theta` .

 = `theta`   `tanh^-1 2theta` `-` 2`int 1/(u) (-1/8)` du

= `theta`   `tanh^-1 2theta` `+`    `(1/4) int 1/(u)` du

 =  `theta`   `tanh^-1 2theta` `+`   ` (1/4) int ` `u^(-1)du`     

=    `theta`   `tanh^-1 2theta` `+` `(1/4) (log u)` + C

 =   `theta`   `tanh^-1 2theta` `+` `1/4 log(1-4theta^2)` + C 

Problems for variable selection in differentiation

• Review on the variable selection with the differentiable function `y = e^x [sin x + cos x]` calculate   `y'' - 2y' + y `         

Solution:

Given     `y = e^x [sin x + cos x]`           

The formula is given as

    `d/dx [f(x). g(x)] = f'(x).g(x) + f(x).g'(x)`      

or      `d/dx [u.v] = u'.v + u.v'`      

Variable Selection:

Here sometime one problem exist, which variable we can take to apply in the formula.

We can take any of them.

Take v = `sinx + cosx`  u =`e^x`     

     `v'`   = `cos x - sinx`   `u'` =`e^x`    

First Differentiation:

      `y' = u'.v + u.v'`

      `y' = e^x [sin x + cos x] + e^x[cos x - sin x]`

      `y' = e^x sin x + e^x cos x + e^x cos x - e^x sin x`    

      `y' = 2 e^x cos x `    

Second  Differentiation:

      `y'' = 2e^x [-sin x] + 2e^x[cos x ]`    

      `y'' = 2e^x cos x - 2e^x sin x`      

  `y'' - 2y' + y `  = ` 2e^x cos x - 2e^x sin x``+` `[-2( 2 e^x cos x)] ` `+ e^x sin x + e^x cos x`             

  `y'' - 2y' + y `  = ` 2e^x cos x - 2e^x sin x``-` `4 e^x cos x ` `+ e^x sin x + e^x cos x`             

  `y'' - 2y' + y `  = ` [2 - 4 + 1]e^x cos x + [1-2]e^x sin x`                    

  `y'' - 2y' + y `  = ` -e^x cos x - e^x sin x`   is the required solution.