What is the derivative of y=arctan(4/x)?

[xxclaymanxx]

derivative of: y=arctan(4/x)!

This seems like it should be a very SIMPLE problem, but it has been bugging me for about 2 hours straight now, and I can't figure it out! Would really appreciate some quick help...

This is what I've done so far:

y'= -4/x^2 / (16/x^2 +1)

I don't know if this is the right answer? And I am having problems simplifying this any further. Please help :P

 

[Dick]

That's fine. Now use algebra. (a/b)/c=a/(b*c).

 

[rock.freak667]

Yes that is correct.But you can simply further by multiply both the numerator and denominator by x^2

EDIT: ahh beaten

 

[Dick]

Yes that is correct.But you can simply further by multiply both the numerator and denominator by x^2

EDIT: ahh beaten

It's not a race. Good advice is best given in large quantities. I made that aphorism up just now. :)

 

[xxclaymanxx]

Ok great thanks,

Now for this next part, I guess the thing that is throwing me off is the +1 hanging off the denominator.

So, following these steps, I would come out with:

-4/16 + x^2

...

which I could then reduce further to:

-1/4+x^2

Would that make sense?

 

[Dick]

You should really review some algebra. Yes, to -4/(16+x^2). A big thumbs down to -1/4+x^2. They aren't at all the same, are they? And use more parentheses to group terms, ok?

 

[xxclaymanxx]

Yah, I think I do need some algebra review. That was a really simple error I just made, and this problem took me WAY too long to figure out.

Thanks a lot for the quick responses...I just signed up for these forums, and hoped that I would be able to get a respone within the hour, but you sure beat that!

Regarding the question, -4/(16+x^2) has been reduced, and I will leave it like this, which I think is the correct answer. So thanks again guys!