$\displaystyle \sqrt{2x^2}$
1. This is a term and not a function, so ...
2. Where are you stuck (and why?)?
3. I assume that you want to differentiate $\displaystyle f(x) = \sqrt{2x^2}$
If so: Re-write the equation of the function:
$\displaystyle f(x) = \sqrt{2x^2}~\implies~f(x) = \left \lbrace \begin{array}{rcl}x \cdot \sqrt{2}& if & x \ge 0 \\ -x \cdot \sqrt{2}& if & x < 0\end{array} \right.$
4. Continue!
This is where I am stuck. I have to find the equation of a line tangent to the curve $\displaystyle f(x) = \frac{\sqrt{2x^3}}{2} $at the point (2,2)
but when I use the quotient rule to find the derivative I get stuck when differentiating $\displaystyle \sqrt{2x^3}$
Why? Does the $\displaystyle x^\frac{1}{2}$ (from simplifying the root) get distributed to $\displaystyle 2$ and $\displaystyle x^3$? If you know a simpler way of doing this please show me.
I don't understand why you want to use the quotient rule because there isn't any variable in the numerator.
Re-write the equation of the function:
$\displaystyle f(x) = \frac{\sqrt{2x^3}}{2} ~\implies~f(x)=\frac12 \cdot \sqrt{2} \cdot x^{\frac32}$
Now apply the power rule to differentiate this function.