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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 $1+x^2=\sec^2(\arctan(x))$?

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Let $x=\tan y$. Then $x^2+1=\sec ^2 y$. But $y=\arctan x$ –  Pedro Tamaroff Aug 25 '12 at 2:52
    
Perfect! Thank you. I would never have guessed that! –  Korgan Rivera Aug 25 '12 at 2:57
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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. –  Ross Millikan Aug 25 '12 at 3:47
    
@PeterTamaroff If your comment was an answer, I would have chosen it. –  Korgan Rivera Aug 25 '12 at 4:48
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4 Answers 4

Use the trigonometric identity $1+\tan^2 x=\sec^2 x$.

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A fancy, longer way:

$$(1+x^2)'=2x$$ $$[\sec^2(\arctan x)]'=2\sec(\arctan x)\cdot\tan(\arctan x)\cdot\sec(\arctan x)\cdot\frac{1}{1+x^2}=$$ $$=2\sec^2(\arctan x)\frac{x}{1+x^2}=2(1+\tan^2(\arctan x))\frac{x}{1+x^2}=$$ $$=2(1+x^2)\frac{x}{1+x^2}=2x$$ Thus, we got: $$(1+x^2)'=[\sec^2(\arctan x)^2]'\Longleftrightarrow 1+x^2=\sec^2(\arctan x)+C\,\,,\,C=\text{ a constant}$$

To find $\,C\,$ now just evaluate above for $\,x=0\,$ , say...

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The following geometric approach can be useful.

First suppose that $x \gt 0$. Draw a right-angled triangle $ABC$, right-angled at $C$. Look at $\angle A$. Let the length of $BC$ be $x$ (put an $x$ beside leg $BC$). Let $AC$ have length $1$. Then $\tan A=x$, so $A=\arctan x$. The hypotenuse has length $\sqrt{1+x^2}$, so $\sec A=\frac{1}{\cos A}=\frac{\sqrt{1+x^2}}{1}$, and therefore $\sec^2 A=1+x^2$.

If $x=0$ the result is trivial.

If $x \lt 0$, note that $\arctan x=-\arctan |x|$.

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Note that $sec^{2}u=1+tan^{2}u$. Let $u=arctan(x)$. That should take care of the equality you want. The usual way to get the derivative of $arctan(x)$, at least from my experience, is to let $y=arctan(x)$ and conclude that $tan(y)=x$, by definition of the $arctan$ function. Now use implicit differentiation. You should get $\frac{dy}{dx}=\frac{1}{sec^{2}y}$.Substitute for $sec^{2}y$ to get $\frac{dy}{dx}=\frac{1}{1+tan^{2}y}$. But $tan(y)=x$, and you get the desired result.

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