Question

How do you find the derivative of `ln(x/(x^2+1))`?

Answer 1

`f'(x)=(1-x^2)/(x^3+x)`

Explanation:

The easiest way is to first rewrite the function using properties of logarithms. Recall that `log(a/b)=log(a)-log(b)`.

`f(x)=ln(x/(x^2+1))`

`f(x)=ln(x)-ln(x^2+1)`

Now we have two simpler functions to differentiate. Recall that `d/dxln(x)=1/x`. The chain rule tells us that the derivative of a function within `ln(x)` is `d/dxln(g(x))=1/(g(x))*g'(x)=(g'(x))/(g(x))`.

Then:

`f'(x)=1/x-(d/dx(x^2+1))/(x^2+1)`

`f'(x)=1/x-(2x)/(x^2+1)`

These can be combined, but it's not necessary:

`f'(x)=(x^2+1-2x(x))/(x(x^2+1))`

`f'(x)=(1-x^2)/(x^3+x)`