To find the derivative of ( y = \log(\arcsin(1-3x)) ), we can use the chain rule.

Let ( u = \arcsin(1-3x) ). Then the derivative of ( u ) with respect to ( x ) is:

[ \frac{du}{dx} = \frac{1}{\sqrt{1 - (1-3x)^2}} \cdot (-3) ]

Next, let ( v = \log(u) ). The derivative of ( v ) with respect to ( u ) is:

[ \frac{dv}{du} = \frac{1}{u} ]

Finally, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = \frac{dv}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot \frac{1}{\sqrt{1 - (1-3x)^2}} \cdot (-3) ]

Therefore, ( y' = -\frac{3}{u \sqrt{1 - (1-3x)^2}} ).