Derivative of arcsin(2x) and evaluation at a point

Recall that $\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}$. Here $u = 2x$, so $u' = 2$. Applying the chain rule gives $h'(x) = \frac{2}{\sqrt{1-(2x)^2}} = \frac{2}{\sqrt{1-4x^2}}$.

Now substitute $x = \frac{1}{4}$: $4x^2 = 4\cdot\frac{1}{16} = \frac{1}{4}$, so $1 - 4x^2 = \frac{3}{4}$. Thus $h'\left(\frac14\right) = \frac{2}{\sqrt{3/4}} = \frac{2}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}}$.

The derivative is positive. We found h prime of x equals two divided by the square root of one minus four x squared, which is always positive wherever it is defined, since it is a positive number divided by a positive square root. Evaluating at one fourth gave four divided by the square root of three, a clearly positive value.