Derivative of a product involving e^(2x) and arcsin(x)

Apply the product rule: $\frac{dy}{dx} = \frac{d}{dx}[e^{2x}] \cdot \arcsin(x) + e^{2x} \cdot \frac{d}{dx}[\arcsin(x)]$. Since $\frac{d}{dx}[e^{2x}] = 2e^{2x}$ (chain rule) and $\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}$, this gives $\frac{dy}{dx} = 2e^{2x}\arcsin(x) + \dfrac{e^{2x}}{\sqrt{1-x^2}}$.

Now substitute $x = 0$: $\arcsin(0) = 0$, $e^{0} = 1$, and $\sqrt{1-0^2} = 1$. So $\frac{dy}{dx}\Big|_{x=0} = 2(1)(0) + \dfrac{1}{1} = 1$.

Arcsine of zero equals zero, so the term two times e to the zero times arcsine of zero vanished completely. This means that entire product contributed zero to the final answer at x equals zero. The only surviving term was e to the zero divided by the square root of one, which equals one. So arcsine of zero contributed nothing, that is, zero.