Help with Derivatives

[Iooy]

ok, got this question and Im puzzled at how it is set up. Perhaps someone else here is able to assist me in figuring out the 1st Dirvitive of it.

f(x) = LN(x + LNx)

Looking for the steps to get the f'(x)

Thank you.

[Edited on 12-13-2005 by Edaarin]

[Latrinsorm]

Let's call f(x) == y to make my typing life easier. Take the exponential of both sides:

e^y = x + ln(x)

Take d/dx of both sides.
e^y * dy/dx = 1 + 1/x

Substitute (x + ln(x)) back in for (e^y) and divide through.

dy/dx = (1 + 1/x) / (x + ln(x))

Alternatively, the derivative rule for ln(u) is 1/u * du/dx, which would get you:

1/(x + ln(x)) * d(x + ln(x))/dx
(1 + 1/x) / (x + ln(x))

[Unique]

Originally posted by Latrinsorm

Alternatively, the derivative rule for ln(u) is 1/u * du/dx, which would get you:

1/(x + ln(x)) * d(x + ln(x))/dx
(1 + 1/x) / (x + ln(x))


AKA chain rule. It's essential.

Unique.

[Edaarin]

Sorry, I had to fix the title, that was driving me insane.