Derivative of square root What would be the derivative of square roots? For example if I have $2 \sqrt{x}$ or $\sqrt{x}$.

I'm unsure how to find the derivative of these and include them especially in something like implicit.
Let $f(x) = \sqrt{x}$, then $$f'(x) = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \dfrac{\sqrt{x+h} - \sqrt{x}}{h} \times \dfrac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}} = \lim_{x \to 0} \dfrac{x+h-x}{h (\sqrt{x+h} + \sqrt{x})}\\ = \lim_{h \to 0} \dfrac{h}{h (\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \dfrac1{(\sqrt{x+h} + \sqrt{x})} = \dfrac1{2\sqrt{x}}$$
In general, you can use the fact that if $f(x) = x^{t}$, then $f'(x) = tx^{t-1}$. 

Taking $t=1/2$, gives us that $f'(x) = \dfrac12 x^{-1/2}$, which is the same as we obtained above.

Also, recall that $\dfrac{d (c f(x))}{dx} = c \dfrac{df(x)}{dx}$. Hence, you can pull out the constant and then differentiate it.
$\sqrt{x}$
Let $f(u)=u^{1/2}$ and $u=x$
That's 
$\frac{df}{du}=\frac{1}{2}u^{-1/2}$ and $\frac{du}{dx}=1$
But, by the chain rule 
$\frac{dy}{dx}=\frac{df}{du}•\frac{du}{dx}
=\frac{1}{2}u^{-1/2} •1
=\frac{d}{dx}\sqrt{x}$
Finally
$\frac{1}{2\sqrt{x}}$