First principle derivative of a square root and conjugates

Question

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.

Answer 1

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.

Answer 2

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, $|x|<1$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(1+\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-\sqrt {x+2}-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}+\sqrt {x+2}\left(\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-\sqrt {x+2}-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\frac{\sqrt {x+2}\left(\frac{1}{2}\left(\frac{h}{x+2}\right)+o(h^2) \right)-6xh-3h^2 }{h}$$ $$=\lim_{h\to 0}\sqrt {x+2}\left(\frac{1}{2}\left(\frac{1}{x+2}\right)+o(h) \right)-6x-3h $$ $$=\sqrt {x+2}\left(\frac{1}{2}\left(\frac{1}{x+2}\right)+0 \right)-6x-3(0) $$ $$=\frac{1}{2\sqrt{x+2}}-6x$$

Answer 3

$$\lim_{h\to0}{\frac{g(x+h)-g(x)}{h}}$$

$$\lim_{h\to0}{\frac{\sqrt{x+h+2}+3(x+h)^2-\sqrt{x+2}-3x^2}{h}=\frac{\sqrt{x+h+2}-\sqrt{x+2}-3x^2-6xh-3h^2+3x^2}{h}}$$
$3x^2$ 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: $$\frac{1}{2\sqrt{x+2}}-6x$$