Derivative of inverse trig function absolute value?

[quark001]

Find the derivative of y = arctan(x^(1/2)).

Using the fact that the derivative of arctanx = 1/(1+x^2) I got:

dy/dx = 1/(1+abs(x)) * (1/2)x^(-1/2)

But my textbook gives it without the absolute value sign. I don't understand why because surely x^(1/2) squared is the absolute value of x and not simply x?

 

[mfb]

x^(1/2) is well-defined for positive x only (unless you work with complex numbers). If you don't have any negative numbers, ...

 

[tade]

you have to consider the signs of the trigo identities

 

[quark001]

Oh okay. The function is undefined for negative x to begin with... Stupid question.