Thread: derivative question (ln)

Hello, I have the equation below which I want to find the derivative of:

y=ln(x^2+1)

I understand that I have to proceed to:
1/x^2+1 x (2x) (using chain rule)

I understand that, but am not clear on which two parts will comprise the chain rule here and how the 'ln' is utilised.
Can someone please show me the distinct steps to get to this stage.

Thanks kindly for any help.

It's actually 2x[1/(x^2 + 1)], you MUST use your brackets correctly, and don't use x to mean both the letter x and times.

Anyway, your function is y = ln(x^2 + 1). If you let u = x^2 + 1 then y = ln(u).

What is du/dx?

What is dy/du?

Once you have those, what is dy/dx?

y is a composite function:

y = fog(x).

f(x) = ln(x)

g(x) = x^2+1

so the deriviative is f'(g(x))(g'(x)).

f'(x) = 1/x, and g'(x) = 2x.

so f'(g(x)) = 1/(g(x)). any clearer now?