Processing Math: Done

Aus Online Mathematik Brückenkurs 2

There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:

ddxlnx+x+1=1x+x+1x+x+1 

We can carry out the differentiation of x+x+1  on the right-hand side term by term to obtain

=1x+x+1x+x+1 

and it remains then only to differentiate x ,which we do directly, and x+1  which a simple inner derivative)

=1x+x+112x+12x+1x+1=1x+x+112x+12x+11

If we rewrite the expression inside the square brackets using a common denominator, we get

=1x+x+12xx+1x+1+x ,

and we can then eliminate the factor x+1+x  from the numerator and denominator to get

=12xx+1