Derivative of inverse trig function absolute value?

  • Thread starter quark001
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  • #1
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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?
 

Answers and Replies

  • #2
35,626
12,162
x^(1/2) is well-defined for positive x only (unless you work with complex numbers). If you don't have any negative numbers, ...
 
  • #3
552
18
you have to consider the signs of the trigo identities
 
  • #4
44
0
Oh okay. The function is undefined for negative x to begin with... Stupid question.
 

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