Derivative of x*sqrt[x/(2-x)] - Solving for the Correct Answer

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The discussion revolves around finding the derivative of the function x*sqrt[x/(2-x)]. Participants compare their answers, with one claiming to have a more simplified version. The consensus is that the original answers provided by the users are incorrect, and they emphasize the need to apply the product, chain, and quotient rules correctly. Ultimately, one participant confirms that their answer matches another's, leading to a resolution of the confusion. The conversation highlights the complexity of the derivative, challenging the notion of it being a "simple derivative."
hanelliot
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Homework Statement


derivative of x*sqrt[x/(2-x)]


Homework Equations





The Attempt at a Solution


my friend got 1/(2-x) and I got {-(x-3)*[-x/(x-2)]^3/2}/x. Who's right?
 
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If I'm reading your problem correct, your answer seems more correct than his.

I've got [(sqrt x)*(3-x)]/[(2-x)^(3/2)] which should be the same answer in a more simplified format.

Now my question is: what kind of class are you taking where you would call this a "simple derivative"? :)
 
I don't think either of you is correct, and your friend is very far off. Your answer doesn't appear to be correct, but at least I can see how you are thinking. To do this problem you will have to apply, in this order,
  1. the product rule,
  2. the chain rule,
  3. the quotient rule
 
mrkuo said:
If I'm reading your problem correct, your answer seems more correct than his.

I've got [(sqrt x)*(3-x)]/[(2-x)^(3/2)] which should be the same answer in a more simplified format.

Now my question is: what kind of class are you taking where you would call this a "simple derivative"? :)
multivariable calculus II, I've been away from math for a long time and this is embarrassing.. anyways, ur answer = mine (plugged in few numbers to confirm), thanks all!
 
I got 1/[(2-x)^2]
 
lmnop said:
I got 1/[(2-x)^2]
First off, this thread is almost 9 months old.

Second, my answer agrees exactly with mrkuo's answer. lmnop, how did you get your answer?
 

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