How do I simplify this: f(x)=log(e) [(x+1) v(x-1)] and find derivative

Question

How do I simplify this: f(x)=log(e) [(x+1) √(x-1)] and find derivative 

Please show step by step solution

Thanks!

Answer 1

log_ex  is also known as lnx

 

So, f(x) = ln[(x + 1)√(x - 1)]

 

              = ln(x + 1) + ln√(x - 1)

 

              = ln(x + 1) + ln(x - 1)^½

 

              = ln(x + 1) + (1/2)ln(x - 1)

 

(lnu)' = u'/u

 

So, f'(x) = 1/(x + 1) + (1/2)(1/(x - 1))

 

                =     1        +        1       

                     x + 1          2(x - 1)

 

                =         2(x - 1)      +        x + 1        

                  2(x - 1)(x + 1)      2(x - 1)(x + 1)

 

                =          3x - 1          

                   2(x - 1) (x + 1)

Answer 2

I assume you were trying to write the following:

 

f(x) = log_e [(x + 1)  √ (x - 1)] which is the same as ln [(x + 1)  √ (x - 1)]

 

By the rules of logarithms, this can be simplified as follows:

 

ln (x + 1) + ln √ (x - 1) =

ln (x + 1) + ln (x - 1) ^1/2 = 

ln (x + 1) + 1/2 ln (x - 1)

 

So the derivative is:

 

(1 / (x + 1)) * D_x (x + 1) + (1/2) (1/(x-1)) D_x (x - 1) =

(1 / (x + 1))* 1 + 1 / (2(x - 1) * 1 =

1/(x + 1) + 1 / (2 (x - 1))