If \(y=\sin^{-1}(3x)\), then \(\dfrac{dy}{dx} =\)
\( \dfrac{3}{1+9x^2} \)
\( \dfrac{-3}{\sqrt{1-9x^2}} \)
\( \dfrac{1}{\sqrt{1-9x^2}} \)
\( \dfrac{3}{\sqrt{1-9x^2}} \)
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The derivative of arcsin:
$$ \frac{d}{dx}[\arcsin u]=\frac{u'}{\sqrt{1-u^2}} $$ $$ \frac{d}{dx}[\arcsin(3x)]=\frac{3}{\sqrt{1-(3x)^2}} $$ $$ = \boxed{\frac{3}{\sqrt{1-9x^2}}} $$