Differentiate f(x)= 6-xe^x/x+e^x

Question

Differentiate f(x)= 6-xe^x/x+e^x

Answer 1

f(x) = (6-e^x)/(x + e^x)

 

f'(x) = e^x(x² + 6) + e^2x + 6

          _________________

                   (x+e^x)²

Answer 2

Assuming exactly the way you typed it in:

 

(I assumed no extra parenthesis and used a strict order of operations.)

 

d/dx [ 6 - (xe^x/x) + e^x ]

 

the expression  [ 6 - (xe^x/x) + e^x ]  =  [ 6 - e^x + e^x ] = 6

 

so d/dx (6) = 0

 

Of course if you really meant it as:

 

f(x) = (6-e^x)/(x + e^x)    then it is 

 

f'(x)^' = [ e^x(x² + 6) + e^2x + 6 ] / (x+e^x)²

  Note: it is f ' (x)  or f(x) ' == f prime(x) == not f(x)

 

If you meant something else ask again letting us know where the parenthesis go.