What is $\frac{d(\arctan(x))}{dx}$? 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!
 A: 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 $\displaystyle\lim_{u\to0}\frac{\arctan u}u=\lim_{v\to0}\frac v{\tan v}=\lim_{v\to0}\cos v\cdot\frac1{\lim_{v\to0}\frac{\sin v}v}=1\cdot1$
A: 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 $$ \dfrac {1}{\left( \sqrt{1+x^2} \right)^2} = \boxed {\dfrac {1}{1+x^2}}. $$
A: $$v=\arctan(x)\Rightarrow x=\tan v\Rightarrow x'=\frac{1}{(\tan v)'}=\frac{1}{(\frac{\sin v}{\cos v})'}=\frac{1}{\frac{1}{\cos^2 v}}=\cos^2 v=\frac{\cos^2 v}{1}$$
$$=\frac{\cos^2 v}{\cos^2 v+\sin^2 v}=\frac{\frac{\cos^2 v}{\cos^2 v}}{\frac{\cos^2 v+\sin^2 v}{\cos^2 v}}=\frac{1}{1+\tan^2 v}=\frac{1}{1+x^2}$$
i.e
$$v'=(\arctan(x))'=\frac{1}{1+x^2}$$
A: 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
$$\frac{\mathrm{d}v}{\mathrm{d}x}=\frac{1}{1+x^2}$$
