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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

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

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    I can simplify it down into $\frac {x^2}{(1+x^2)^(.5)}$2012-04-21

2 Answers 2

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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.

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If I am not wrong the derivative should be: $\frac{-\frac{1}{2}(1+x^2)^{-1/2}2x\cdot x+(1+x^2)^{1/2}}{x^2}=\frac{-(1+x^2)^{-1/2}(x^2+1-1)+(1+x^2)^{1/2}}{x^2}=\frac{(1+x^2)^{-1/2}}{x^2}$