The question is to find the derivative of y = x times the square root of (3x^2 + 1)

I rewrite this as y = x(3x^2 + 1)^1/2

Then I find y' = x (1/2)(3x^2 + 1)^-1/2 (6x) + 1(3x^2 + 1)1/2

Is this right and is there any way to further simply this?

Yes, you can. That is where algebra comes in.

You have \frac{3x^{2}}{\sqrt{3x^{2}+1}}+\sqrt{3x^{2}+1}, which is correct and well done.

You can leave it like this or get a common denominator, like so:

Multiply top and bottom of the right side by \sqrt{3x^{2}+1}


\frac{3x^{2}}{\sqrt{3x^{2}+1}}+\frac{\sqrt{3x^{2}+ 1}}{1}\cdot \frac{\sqrt{3x^{2}+1}}{\sqrt{3x^{2}+1}}

Now, the denominators are the same. As when adding any fraction.

\frac{6x^{2}+1}{\sqrt{3x^{2}+1}}