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.
Best Answer
Starting from where you got stuck, first split up the fraction as: $$\frac{\sqrt {x+h+2}-\sqrt {x+2}}{h} - \dfrac{(3x^2+6xh+3h^2)-3x^2}{h}$$
For the first fraction, multiply the top and the bottom by the conjugate. For the second fraction cancel the $3x^2$ terms and factor: $$\frac{(\sqrt {x+h+2}-\sqrt {x+2})(\sqrt {x+h+2}+\sqrt {x+2})}{h(\sqrt {x+h+2}+\sqrt {x+2})} - \dfrac{(6x+3h)h}{h}$$
Now, multiply out the numerator of the first fraction, simplify both fractions, and take the limit as $h \to 0$ to get the answer.