If you don't know the derivative of the inverse of sine, you can use implicit differentiation. Apply sine to both sides:
[tex]y=\sin^{-1}(6x+1)\implies\sin y=6x+1[/tex]
(true for y between -π/2 and π/2)
Now take the derivative of both sides and solve for it:
[tex]\cos y\dfrac{\mathrm dy}{\mathrm dx}=6[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=6\sec y[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\cos\left(\sin^{-1}(6x+1)\right)}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac6{\sqrt{1-(6x+1)^2}}[/tex]
