find derivate of root (2x - 1) using first principles, Thanks for your help.

This is fairly straightforward. You should get, as your limit:

 

[math]\lim_{h \to 0} \frac{\sqrt{2x-1}-\sqrt{2(x+h) - 1}}{h}[/math]

 

Now, by multiplying by the conjugate of the top, we get:

 

[math]\lim_{h \to 0} \frac{-2h}{h(\sqrt{2x-1}+\sqrt{2(x+h) - 1})}[/math]

 

This should give you your answer.

The answer to the problem is: 1/root(2x-1), but when I solve Dave's equation above, I get -1/root(2x-1)....?

 

d/dx(root(f(x))) = f'(x)/2root(f(x)) isn't it?

Very sorry, I got my limit the wrong way around. It should be:

 

[math]\lim_{h \to 0} \frac{\sqrt{2(x+h) - 1} - \sqrt{2x-1}}{h}[/math]

 

This gives the correct answer of [math]\frac{1}{\sqrt{2x-1}}[/math].