First principle derivative of a square root and conjugates I'm trying to find the derivative of the equation: $$g(x)=\sqrt {x+2}-3x^2$$.
I can find the solution just fine using the power rule but am finding trouble with First Principles.
Essentially, I understand getting as far as $$\displaystyle\lim_{h\to 0}\frac{\sqrt {x+h+2}-3(x+h)^2 -\sqrt {x+2}+3x^2}{h}.$$ From here I can expand out to $$\lim_{h\to 0}\frac{\sqrt {x+h+2}-3x^2-6xh-3h^2 -\sqrt {x+2}+3x^2}{h}.$$ But then I get stuck.
I'm not sure if I should use the conjugate rule now (but then how would I even apply that?) or if I'm supposed to try and simplify.
The answer is $\dfrac{1}{2\sqrt {x+2}}-6x$ that I got using the power rule.
Any help and guidance is appreciated.

$$\lim_{h\to0}{\frac{g(x+h)-g(x)}{h}}$$
$$\lim_{h\to0}{\frac{\sqrt{x+h+2}+3(x+h)^2-\sqrt{x+2}-3x^2}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-3x^2-6xh-3h^2+3x^2}{h}}$$

$3x^2$ is eliminated, and $3h^2$ is negligible: 
$$=\frac{\sqrt{x+h+2}-\sqrt{x+2}-6xh}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}-\frac{6xh}{h}$$
now using the conjugate rule
$$=\frac{(\sqrt{x+h+2}-\sqrt{x+2})(\sqrt{x+h+2}+\sqrt{x+2})}{h(\sqrt{x+h+2}+\sqrt{x+2})}-6x=\frac{(x+h+2)-(x+2)}{h(\sqrt{x+h+2}+\sqrt{x+2})}-6x=\frac{h}{h(\sqrt{x+h+2}+\sqrt{x+2})}-6x$$
$$=\frac{1}{(\sqrt{x+h+2}+\sqrt{x+2})}-6x$$
 and since $h\to0$ we have:
$$\frac{1}{2\sqrt{x+2}}-6x$$
