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} $

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".
