Let f(x) = x - ln(x^2+1)
We have :
The derivative will be:
f(x)' = x' - [ ln(x^2 +1)] '
The derivative of x = 1
f(x)' = 1 - [ ln(x^2 +1)]'
Let's derive ln (x^2+1) separately
Let g(x) = ln (x^2+1)
We have to usar the rule of jail
g(x)' = ln(u)' . u'
Where, u = x^2 +1
u' = 2x + 0
u' = 2x
And,
Derivative of ln(u) = 1 / u
Then,
g(x) = ( 1 / u ) . 2x
g(x) = (1 / x^2+1) . 2x
g(x) = 2x / (x^2 +1)
Then,
f(x)' = 1 - 2x / (x^2+1)
Hope this helps
