What is the derivative of x-ln(x^2 +1)?

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