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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+1(x)+(x+1)

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

=1x+x+112x+12x+1(x+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.