Inverse Trig function derivative

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EvilBunny
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Homework Statement



Let

arctan ([tex]\sqrt{3x^2 -1}[/tex])


then dy/dx


Well I know that the derivative of arctanx is

1/ 1 + x ² but when I got something other then simply x I don't know how to proceed
 
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x=the radical

example

[tex]y=\arctan{(x^2)}[/tex]

[tex]y'=\frac{2x}{1+(x^2)^2}[/tex]
 
Neat I get it, thanks.