# How do I differentiate a square root?

I have a problem which I can not seem to fathom. I am asked to differentiate [(sqrt x) + 1]^2 / x with respect to x. I tackled the problem as follows:
First I expanded the top term to become [x + 2(sqrt x) + 1]. Then I multiplied by x^-1 to get 1 + 2(sqrt x)^-1 + x^-1.
When I differentiated, I got -x^-3/2 - x^-2, but the answer sheet says 1 - x^-3/2 - 2x^-3. I can not see where I am going wrong. Any help would be appreciated. Cheers. Adrian.

 ebaines Posts: 10,583, Reputation: 5794 Expert #2 Mar 19, 2012, 11:48 AM
You are correct. Another way to do this is to use the formula for derivate of a fraction:

$
(\frac {f(x)}{g(x)} )' = \frac { g(x) f'(x) - f(x)g'(x)} {g(x))^2}

$

For this problem this gives:

$
\frac {2x(\sqrt x + 1) \frac 1 2 x^{-1/2} - (\sqrt x +1)^2} {x^2} = \frac {x + \sqrt x -(x + 2 \sqrt x +1)} {x^2} = -x^{-3/2} -x^{-2}
$

which is the same answer you got.

However, if you integrate the answer that the answer sheet gives it turns out to be the derivative of (x sqrt(x) +1)^2/x^2. So could there have been a mistake in the original question as you typed it on this site?
 AdrianCavinder Posts: 43, Reputation: 10 Junior Member #3 Mar 22, 2012, 09:18 PM
Dear Uber, Thanks so much. Very helpful.