Help differentiating a trig function with the chain rule? (C4)

I was doing a past paper and differentiated -ln(cos(x)) using the chain rule:

u = cos x
du/dx = -sin(x)

y= -ln(u)
dy/dx = -1/(u)

and therefore dy/dx = -1/(-sin(x) =1/sin(x) or cosec X.

However, I have just checked my answer on wolfram alpha and it said the answer is tan x.

Normally I'd just assume it was wrong, but this is wolfram alpha and I've seen it check far more complex things without an issue. Have I made a stupid mistake, or what? I'm kind of confused as to where the tan x could possibly have come from, but I'm not confident enough to just dismiss it.....

When you differentiate -lnU you get -1/U.
You take U as CosX differentiate it to get -sinX.

Multiply -sinx by -1/U as its the chain rule.
You get sinx/U.
Put Cosx back in and you get sinx/cosx which makes tanX.

Hope this helped!

Ahh! I am such an idiot sometimes. I'll just blame it on.... that long vector question
yes, that must be it.........................

They get me too dw i hate em! Any tips for vectors haha.

I think it's a matter of practicing each type of question enough so you know if you're supposed to be equalling two vectors, making an equation, etc. Once you work out what you're supposed to be trying to achieve they get a bit easier, haha.