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Let $v= \arctan{x}$. Now I want to find $\frac{dv}{dx}$. My method is this: Rearranging yields $\tan(v) = x$ and so $dx = \sec^2(v)dv$. How do I simplify from here? Of course I could do something like $dx = \sec^2(\arctan(x))dv$ so that $\frac{dv}{dx} = \cos^2(\arctan(x))$ but I am sure a better expression exists. I am probably just missing some crucial step where we convert one of the trigonometric expressions into an expression involving $x$. Thanks in advance for any help or tips!

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The derivative of $\tan v$ is $1+\tan^2 v$. It will be easier to simplify, since here $v=\arctan x$.

You may check:

$$\sec^2 v = \frac{1}{\cos^2 v} = \frac{\cos^2 v + \sin^2 v}{\cos^2 v} = 1+\tan^2 v$$

Then

$$\mathrm{d}x = (1+\tan^2 v) \ \mathrm{d}v = (1+x^2) \ \mathrm{d}v$$

And

$$\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{1}{1+x^2}$$

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  • $\begingroup$ Perfect thanks a lot, this is very clear. I see now that using $\cos^2(x) = \frac{1}{\tan(x)^2+1}$ will also change the expression $\cos^2(\arctan(x))$ into $\frac{1}{1+x^2}$. Thanks for the answer! $\endgroup$ – Slugger Nov 28 '13 at 15:24
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Another way :

$$\frac{d\arctan x}{dx}=\lim_{h\to0}\frac{\arctan(x+h)-\arctan x}h$$

$$\displaystyle=\lim_{h\to0}\frac{\arctan\frac{x+h-x}{1+(x+h)x}}h$$

$$\displaystyle=\lim_{h\to0}\left(\frac{\arctan\frac h{1+(x+h)x}}{\frac h{1+(x+h)x}}\right)\cdot\frac1{\lim_{h\to0}\{1+(x+h)x\}}=1\cdot\frac1{1+x^2}$$

as $\displaystyle\lim_{u\to0}\frac{\arctan u}u=\lim_{v\to0}\frac v{\tan v}=\lim_{v\to0}\cos v\cdot\frac1{\lim_{v\to0}\frac{\sin v}v}=1\cdot1$

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    $\begingroup$ Definition of the derivative! +1. $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 15:30
  • $\begingroup$ Thank you for your answer! It is nice to see how the answer tackle my question in a multitude of ways! $\endgroup$ – Slugger Nov 28 '13 at 15:34
  • $\begingroup$ @Slugger, my pleasure. Please have a look into math.stackexchange.com/questions/579170/… $\endgroup$ – lab bhattacharjee Nov 28 '13 at 15:46
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You can also use the Inverse Derivative Formula, which states that if $f(x)$ and $g(x)$ are inverse functions, we have $$ g'(x) = \dfrac {1}{f'(g(x))}. $$So, if $g(x)=\arctan x$, our task is to find $g'(x)$. In that case, we have $f(x)=\tan x$, which gives us $f'(x)=sec^2 x$, so we can substitute: $$ \begin {align*} g'(x) &= \dfrac {1}{f'(g(x))} \\&= \dfrac {1}{\sec^2 (g(x))} \\&= \dfrac {1}{\sec^2 (\arctan x)}. \end {align*} $$We can find $ \sec (\arctan x) $ geometrically. Consider a right triangle with legs of length $x$, $1$, and $\sqrt{1+x^2}$. Let $\theta$ be the angle opposite to the leg of length $x$. Then, $$ \sec \left( \arctan x \right) = \sec (\theta) = \sqrt {1+x^2}, $$ so our answer is $$ \dfrac {1}{\left( \sqrt{1+x^2} \right)^2} = \boxed {\dfrac {1}{1+x^2}}. $$

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  • $\begingroup$ Thanks you for your answer! $\endgroup$ – Slugger Nov 28 '13 at 15:35
  • $\begingroup$ You're welcome! :) $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 15:36
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$$v=\arctan(x)\Rightarrow x=\tan v\Rightarrow x'=\frac{1}{(\tan v)'}=\frac{1}{(\frac{\sin v}{\cos v})'}=\frac{1}{\frac{1}{\cos^2 v}}=\cos^2 v=\frac{\cos^2 v}{1}$$ $$=\frac{\cos^2 v}{\cos^2 v+\sin^2 v}=\frac{\frac{\cos^2 v}{\cos^2 v}}{\frac{\cos^2 v+\sin^2 v}{\cos^2 v}}=\frac{1}{1+\tan^2 v}=\frac{1}{1+x^2}$$ i.e $$v'=(\arctan(x))'=\frac{1}{1+x^2}$$

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  • $\begingroup$ I am not sure I follow exactly. Your second equation says $x=\tan v$ and then you follow by saying $x' =\frac{1}{(\tan v)'}$... Maybe I am missing something $\endgroup$ – Slugger Nov 28 '13 at 15:27
  • $\begingroup$ Sir, my solution is correct, You can also use the Inverse Derivative Formula, $\endgroup$ – Madrit Zhaku Nov 28 '13 at 15:31
  • $\begingroup$ Oh, it seems this is essentially what I did, but I posted a few minutes later. I didn't see your post when I posted, so it's not that I copied. Should I delete my post? $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 15:32
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    $\begingroup$ @MadritZhaku When somebody asks you to explain what you did, "it is correct" is not as helpful as actually explaining what you did. If you don't have the patience to elaborate, don't respond at all. Your comment came off quite rude. $\endgroup$ – Ahaan S. Rungta Nov 28 '13 at 15:33
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    $\begingroup$ all is well that ends well :) $\endgroup$ – Slugger Nov 28 '13 at 15:37

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