Find the derivative of the function. \[ g(x)=\frac{\arcsin (2 x)}{x} \]

To find the derivative of the function \( g(x) = \frac{\arcsin(2x)}{x} \), we will apply the quotient rule. The quotient rule states that if you have a function \( h(x) = \frac{u(x)}{v(x)} \), then the derivative \( h'(x) \) is given by: \[ h'(x) = \frac{u'v - uv'}{v^2} \] In this case, let \( u(x) = \arcsin(2x) \) and \( v(x) = x \). First, we need to find the derivatives \( u'(x) \) and \( v'(x) \): 1. The derivative \( u'(x) \) can be found using the chain rule: \[ u'(x) = \frac{1}{\sqrt{1 - (2x)^2}} \cdot 2 = \frac{2}{\sqrt{1 - 4x^2}} \] 2. The derivative of \( v(x) \) is straightforward: \[ v'(x) = 1 \] Now we can apply the quotient rule: \[ g'(x) = \frac{u'v - uv'}{v^2} = \frac{\left(\frac{2}{\sqrt{1 - 4x^2}}\right) \cdot x - \arcsin(2x) \cdot 1}{x^2} \] Simplifying this gives: \[ g'(x) = \frac{\frac{2x}{\sqrt{1 - 4x^2}} - \arcsin(2x)}{x^2} \] Thus, the derivative of the function \( g(x) = \frac{\arcsin(2x)}{x} \) is: \[ g'(x) = \frac{2x}{x^2\sqrt{1 - 4x^2}} - \frac{\arcsin(2x)}{x^2} \] And here is your final answer for the derivative of the given function!