If $y=\sin^{-1}(3x)$, then $\dfrac{dy}{dx} =$

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The derivative of arcsin:

$$\frac{d}{dx}[\arcsin u]=\frac{u'}{\sqrt{1-u^2}}$$ $$\frac{d}{dx}[\arcsin(3x)]=\frac{3}{\sqrt{1-(3x)^2}}$$ $$= \boxed{\frac{3}{\sqrt{1-9x^2}}}$$