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$ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ Find the derivative

How do I tackle this? My answer is totally different from the correction model, but I have tried for half an hour to show my answer in lateX but I don't know how to, it's too complicated, so, can someone please give a step to step of the solution? the correction model's solution is $ \dfrac {2-3x^2}{\sqrt{x} (x^2+2)^2} $

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Possibly if you rationalize the numerator in your answer, then your answer is the same as the one you gave here. –  Michael Hardy Oct 2 '12 at 16:12
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This isn't actually a differential equation, since it doesn't ask you to find an unknown function given an equation involving the function and its derivatives. –  Michael Hardy Oct 2 '12 at 16:13
    
But I have no clue how to get to this answer, I need some help, our text book is notoriously bad. –  JohnPhteven Oct 2 '12 at 16:15

2 Answers 2

up vote 5 down vote accepted

Use the quotient rule: \begin{align} h'(x) & = \frac{(x^2+2)\frac{d}{dx}(2\sqrt{x}) - 2\sqrt{x}\frac{d}{dx}(x^2+2)}{(x^2+2)^2} \\[10pt] & = \frac{(x^2+2)\frac{1}{\sqrt{x}}-2\sqrt{x}\cdot 2x}{(x^2+2)^2}. \end{align}

Now clear out fractions by multiplying the top and bottom both by $\sqrt{x}$: $$ \frac{(x^2+2)-2x\cdot2x}{\sqrt{x}(x^2+2)^2}. $$

Finally, do the routine simplifications of the numerator and you get exactly what you say is "the correction model's solution".

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Alternatively, whenever you have a quotient, you can turn it into a product. This doesn't always make things easier, but sometimes it does. You be the judge in this case. For example

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

So, now we can do the product rule to find the derivative

$$\begin{align*} h'(x) &= 2\sqrt{x} \frac{d}{dx}(x^2 + 2)^{-1} + (x^2 + 2)^{-1} \frac{d}{dx} (2\sqrt{x}) \\ &= 2\sqrt{x} (-1)(x^2 + 2)^{-2}(2x) + (x^2 + 2)^{-1} x^{-1/2} \\ &= \frac{-4x^{3/2} + x^{3/2} + 2x^{-1/2}}{(x^2+2)^2} \\ &= \frac{2 - 3x^2}{\sqrt{x}(x^2+2)^2} \end{align*} $$

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