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$\begingroup$

So I figured I can use the chain rule to do this:

$g\prime(x)=\frac{1}{f^\prime(g(x))}$

So that

$(\arctan(x))\prime = \frac{1}{\left[\sec^2(\arctan(x)){}\right]^\prime}$

But this book tells me that

$(\arctan(x))\prime = \frac{1}{x^2+1}$

So, how do I show that $1+x^2=\sec^2(\arctan(x))$?

  • 2
    Let $x=\tan y$. Then $x^2+1=\sec ^2 y$. But $y=\arctan x$2012-08-25
  • 0
    Perfect! Thank you. I would never have guessed that!2012-08-25
  • 1
    Every time I see trig function(arc some other trig function) I draw a right triangle and label the sides to make the other trig function right and the Pythagorean theorem solves my problem.2012-08-25
  • 0
    @PeterTamaroff If your comment was an answer, I would have chosen it.2012-08-25

4 Answers 4