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} $

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