Question

Use the limit definition of the derivative:
y = sqrt(3 x)+1

Answer 1

Okay:
The limit definition of a derivative is:
\[\lim_{\Phi \rightarrow 0}\frac{f(x+\Phi)-f(x)}{\Phi}\]
So:
\[f(x+\Phi)=\sqrt{3x+3\Phi}\]
Now you have:
\[\lim_{\Phi \rightarrow 0}\frac{\sqrt{3x+3\Phi}-\sqrt{3x}}{\Phi}\]
(From here on out there is an implied limit in the front (saves time typing it over and over).
Multiply top and bottom by the conjugate of the top:
\[\frac{(\sqrt{3x+3\Phi}-\sqrt{3x})(\sqrt{3x+3\Phi}+\sqrt{3x})}{\Phi(\sqrt{3x+3\Phi}+\sqrt{3x})}\]
Giving:
\[\frac{3x+3\Phi-3x}{\Phi(\sqrt{3x+3\Phi}+\sqrt{3x})}=\frac{3\Phi}{\Phi(\sqrt{3x+3\Phi}+\sqrt{3x})}=\frac{3}{\sqrt{3x+3\Phi}+\sqrt{3x}}\]
Taking the limit we have:
\[\lim_{\Phi \rightarrow 0}\frac{3}{\sqrt{3x+3\Phi}+\sqrt{3x}}=\frac{3}{\sqrt{3x}+\sqrt{3x}}=\frac{3}{2\sqrt{3x}}\]

Answer 2

What happens to the 1?

Answer 3

Well when you have:
\[\sqrt{3x+3\Phi}+1-(\sqrt{3x}+1) \implies \sqrt{3x+3\Phi}+\sqrt{3x}\]
The ones cancel out. I jumped ahead of myself in the algebra. Sorry.