<s>[INST] 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.
 [/INST] I am taking from the step you got stuck $$\lim_{h\to 0}\frac{\sqrt {x+h+2}-3x^2-6xh-3h^2 -\sqrt {x+2}+3x^2}{h}$$  $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{h}{x+2}\right)^{1/2}-\sqrt {x+2}-6xh-3h^2 }{h}$$ Use binomial expansion $(1+x)^n=1+nx+ \ldots $ where, $|x|<1$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-\sqrt {x+2}-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}+\sqrt {x+2}\left(\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-\sqrt {x+2}-6xh-3h^2 }{h}$$  $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-6xh-3h^2 }{h}$$  $$=\lim_{h\to 0}\sqrt {x+2}\left(\frac{1}{2}\left(\frac{1}{x+2}\right)+o(h) \right)-6x-3h $$  $$=\sqrt {x+2}\left(\frac{1}{2}\left(\frac{1}{x+2}\right)+0 \right)-6x-3(0) $$  $$=\frac{1}{2\sqrt{x+2}}-6x$$
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