Using the limit definition to find a derivative for $2\sqrt{x-3}$

I am trying to find the derivative of $2\sqrt{x-3}$ using the limit definition of a derivative.

What I did is $$\lim_{h \to 0} \frac {2 \sqrt {(x+h)-3} -2 \sqrt {x-3)})}{h}$$

I multiplied the denominator and numerator by $$2 \sqrt {(x+h)-3}+ 2 \sqrt {x-3}$$

After simplifying I got $$\frac {4}{ 2 \sqrt {(x+h)-3}+2 \sqrt {x-3}}$$

But I am having problem finding the correct answer.

1 Answer

$\cfrac{(2\sqrt{x+h}-3)-(2\sqrt{x}-3)}{h}=2\cfrac{\sqrt{x+h}-\sqrt{x}}{h}=2\cfrac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}=2\cfrac{(x+h)-(x)}{h(\sqrt{x+h}+\sqrt{x})}=2\cfrac{h}{h(\sqrt{x+h}+\sqrt{x})}=2\cfrac{1}{\sqrt{x+h}+\sqrt{x}}\xrightarrow{\scriptscriptstyle h\to0} 2\cfrac{1}{\sqrt{x}+\sqrt{x}}=\cfrac{1}{\sqrt{x}}$