<s>[INST] Prove that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$ Prove that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$
Note 2 things here please: I KNOW the differentiation rules, I KNOW  that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$, but I'm having a trouble proving this with a certain method, you see I was training on the basic method -I don't really know what they call it- to get back to the origins of calculus, which is $\frac{d}{dx}f(x) = \lim_{h \to 0}\frac {f(x+h)-f(x)}{h}$
I've done the following:
$$\lim_{h \to 0}\frac {3(x+h)^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}$$
which is so far correct, but I'm struggling to transform this to $6x+\frac{2}{\sqrt{x}}$, I know the problem isn't hard but I wish that someone will be kind and tell me what the next step is.
 [/INST] You made a mistake. There was a $- 4\sqrt{x}$ not $+4\sqrt{x}$.
$$\begin{aligned}\lim_{h \to 0}\frac {3(x+h)^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}& = \lim_{h \to 0}\frac {3x^2 + 6xh+3h^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}\\& = \lim_{h \to 0}\frac {6xh+3h^2+4\sqrt{x+h}-4\sqrt{x}}{h}\\&=   \lim_{h \to 0}\frac {6xh+3h^2}{h}+\frac{4\sqrt{x+h}-4\sqrt{x}}{h}\\& =   \lim_{h \to 0}{6x+3h}+\lim_{h\to 0}\frac{4(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+ \sqrt{x})}{h\sqrt{x+h}+ \sqrt{x})}\\&= 6x+\lim_{h\to 0}\frac{4(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+ \sqrt{x})}{h(\sqrt{x+h}+ \sqrt{x})}\\& =  6x+\lim_{h\to 0}\frac{4}{(\sqrt{x+h}+ \sqrt{x})}\\& =   6x+\frac{2}{\sqrt{x}}\end{aligned}$$
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