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.
 A: The Power Rule says that $\frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}$. Applying this to $\sqrt{x}=x^{\frac12}$ gives
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}
&=\frac{\mathrm{d}}{\mathrm{d}x}x^{\frac12}\\
&=\frac12x^{-\frac12}\\
&=\frac{1}{2\sqrt{x}}\tag{1}
\end{align}
$$
However, if you are uncomfortable applying the Power Rule to a fractional power, consider applying implicit differentiation to
$$
\begin{align}
y&=\sqrt{x}\\
y^2&=x\\
2y\frac{\mathrm{d}y}{\mathrm{d}x}&=1\\
\frac{\mathrm{d}y}{\mathrm{d}x}&=\frac{1}{2y}\\
&=\frac{1}{2\sqrt{x}}\tag{2}
\end{align}
$$
A: 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.
A: Let $f(x) =  \sqrt{x} = x^{1/2}$.
$$f'(x) = \frac{1}{2} x ^{-1/2}$$
$$f'(x) = \frac{1}{2x^{1/2}} = \frac{1}{2\sqrt{x}}$$
If you post the specific implicit differentiation problem, it may help. The general guideline of writing the square root as a fractional power and then using the power and chain rule appropriately should be fine however. Also, remember that you can simply pull out a constant when dealing with derivatives - see below.
If $g(x) = 2\sqrt{x} = 2x^{1/2}$. Then,
$$g'(x) = 2\cdot\frac{1}{2}x^{-1/2}$$
$$g'(x) = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$$
A: $\sqrt x=x^{1/2}$, so you just use the power rule: the derivative is $\frac12x^{-1/2}$.
A: Use the product rule $(fg)'=f'g+fg'$ and take $f=\sqrt{x}=g$. Then
$$x'=1=(\sqrt{x} \cdot \sqrt{x})'=(\sqrt{x})'\sqrt{x}+\sqrt{x}(\sqrt{x})'=2\sqrt{x}(\sqrt{x})'.$$
It follows that $$(\sqrt{x})'=\frac{1}{2\sqrt{x}}$$
A: Another possibility to find the derivative of $f(x)=\sqrt x$ is to use geometry. Imagine a square with side length $\sqrt x$. Then the area of the square is $x$. Now, let's extend the square on both sides by a small amount, $d\sqrt x$. The new area added to the square is:
$$dx=d\sqrt x * \sqrt x + d\sqrt x * \sqrt x + d\sqrt x^2.$$

This is the sum of the sub-areas added on each side of the square (the orange areas in the picture above). The last term in the equation above is very small and can be neglected. Thus:
$$dx=2*d\sqrt x * \sqrt x$$
$$\frac{dx}{d\sqrt x}=2 * \sqrt x$$
$$\frac{d\sqrt x}{dx}=\frac{1}{2*\sqrt x}$$
(To go from the second step to the last, flip the fractions on both sides of the equation.)
Reference: Essence of Calculus, Chapter 3
A: $\sqrt{x}$
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
$\frac{1}{2\sqrt{x}}$
