# Finding the derivative of a square root

I am trying to find the derivative of a pretty simple problem but I just can not force the answer to match the one provided by the book.

$- \frac {(1+x^2)^\frac{1}{2}}{x}$

I mean it is a very simple problem and I get

$\frac {x^2 (1+x^2)^{-.5} - (1+x^2)^.5} {x^2}$

-
You seem to be missing a factor of $\frac{1}{2}$, a $2x$ from the chain rule in the first term in the denominator, and the first factor of the first term in the denominator should be $x$, not $x^2$. I didn't check your signs. –  Neal Apr 21 '12 at 16:00
@Neal I have 1/2 but it gets cancelled out from the 2x. The denominator should be $x^2$ –  user138246 Apr 21 '12 at 16:01
If the denominator of the question is $x^2$, the denominator of the answer should be $x^4$. –  Neal Apr 21 '12 at 16:03
You are fine, except you need to put a negative sign in front of the whole thing (or multiply the numerator by $-1$). –  Joe Johnson 126 Apr 21 '12 at 16:04
@Neal I meant that the denominator is correct in that x is the original and the derivative is x^2 –  user138246 Apr 21 '12 at 16:04

You have $f(x) = -\frac{g(x)}{h(x)}$ where $g(x) = (1+x^2)^{1/2}$ and $h(x) = x$. So you have
\begin{align} g'(x) &= \frac{1}{2}(1 + x^2)^{-1/2}\frac{d}{dx}(1 + x^2) = \frac{1}{2}(1 + x^2)^{-1/2}2x \\ h'(x) &= 1 \end{align} Then just apply the quotient rule (as it looks like you are doing): $$f'(x) = -\frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} = ...$$ EDIT: Note that you could also first rewrite your expression like: \begin{align} f(x) &= - \frac{(1+x^2)^{1/2}}{x} \\ &= -\left(\frac{1}{x^2}(1 + x^2)\right)^{1/2} \\ &= -(x^{-2} + 1)^{1/2} \end{align} And then you wouldn't have to use the quotient rule, but could just use the chain rule.
If I am not wrong the derivative should be: $$\frac{-\frac{1}{2}(1+x^2)^{-1/2}2x\cdot x+(1+x^2)^{1/2}}{x^2}=\frac{-(1+x^2)^{-1/2}(x^2+1-1)+(1+x^2)^{1/2}}{x^2}=\frac{(1+x^2)^{-1/2}}{x^2}$$