Finding the derivative of $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ $ 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} $
 A: 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".
A: 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
$$\begin{align*}
  h'(x) &= 2\sqrt{x} \frac{d}{dx}(x^2 + 2)^{-1} + (x^2 + 2)^{-1} \frac{d}{dx} (2\sqrt{x}) \\
  &= 2\sqrt{x} (-1)(x^2 + 2)^{-2}(2x) + (x^2 + 2)^{-1} x^{-1/2} \\
  &= \frac{-4x^{3/2} + x^{3/2} + 2x^{-1/2}}{(x^2+2)^2} \\
  &= \frac{2 - 3x^2}{\sqrt{x}(x^2+2)^2} 
\end{align*}
$$
