Help w/ Calc 1 derivative (simple)

[Null_]

Homework Statement

f(x) = (x) / (x + (c/x) )

Homework Equations

Product rule and quotient rule

The Attempt at a Solution

f'(x) = [ (x + (c/x)) - (x(1-(c/x^2)) ] / [ (s + (c/x)^2) ]
= [x + (c/x) - x + (c/x)] / [(x+(c/x)^2)]
= [2c/x] / [(x + (c/x)^2)]

That's where I'm stuck. I need to brush up on algebra, I know. If anyone can give me any good review sites, that'd be great.

I'm just doing problems in the book in my own time, and this is an odd number. The answer is:
[2cx] / [ (9x^2)+c)^2 ]

Was I even going along in the right direction? I also tried doing it as simply a product rule, where I just made it cx^-1, but that got messy when I tried to simplify.

 

[Dick]

You've got some typos in there, but if you meant your answer to be (2c/x)/(x+c/x)^2 then that looks correct. You should be able to simplify that to 2cx/(x^2+c)^2. You won't get 2cx/(9*x^2+c)^2. That doesn't look right at all.

 

[ahsanxr]

There seems to be a problem with the answer. There shouldn't be a 9.

I'll continue after where you got stuck:

(2c/x)/(x + c/x)^2

= (2c/x)/[((x^2 + c)/x)^2] (Making the common denominator x of the denominator)
= (2c/x)/[((x^2 + c)^2)/x^2] (Splitting up the square between the numerator and denominator of the denominator)

Then if you remember dividing fractions, then you know that all it is is multiplying one fraction with the reciprocal of the other fraction. Do the same thing here and you'll get:

2cx/(x^2 + c)^2

 

[Null_]

Yeah, that's what I meant. I guess my finger slipped off the shift key.

I got the simplification. I can't believe I forgot the old flip trick to turn division into multiplication. Thanks to both of you.