Derivative of Arctangent with definition of derivative I can't find the derivative of arctangent with definition of derivative.
Here's my way:
$\displaystyle\lim_{\Delta x\to\ 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$ is a definition of derivative with limit. If we define $f(x)=\arctan(x)$, then we get:
$$\left[\lim_{\Delta x\to\ 0} \frac{\arctan(x+\Delta x)-\arctan(x)}{\Delta x}\right]=\left[\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x(x+\Delta x)}\right)}{\Delta x}\right].$$ But I don't know how to continue. Thanks.
 A: $$\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x(x+\Delta x)}\right)}{\Delta x}=\lim_{\Delta x\to\ 0} \frac{\arctan\left(\dfrac{\Delta x}{1+x^2}\right)}{\Delta x}$$
Now let's introduce the variable $u$ by $$\tan u=\dfrac{\Delta x}{1+x^2}$$
Then $$\Delta x=(1+x^2)\tan u$$and your limit is $$\lim_{u\to\ 0}\frac{\arctan(\tan u)}{(1+x^2)\tan u}=\lim_{u\to\ 0}\frac{u}{(1+x^2)\tan u}=\frac1{1+x^2}\frac1{\lim_{u\to\ 0}\frac{\tan u} u}=\frac1{1+x^2}$$
A: Let $y=\arctan x,\,y+\Delta y=\arctan(x+\Delta x)$ so$$\begin{align}\arctan^\prime x&=\lim_{\Delta x\to0}\frac{\arctan (x+\Delta x)-\arctan x}{\Delta x}\\&=\lim_{\Delta x\to0}\frac{\Delta y}{\tan(y+\Delta y)-\tan y}\\&\stackrel{\star}{=}\lim_{\Delta y\to0}\frac{\Delta y}{\tan(y+\Delta y)-\tan y}\\&\stackrel{\dagger}{=}\frac{1}{1+\tan^2y}\\&=\frac{1}{1+x^2},\end{align}$$where $\star$ uses the continuity of $\arctan x$ and $\dagger$ uses $\frac{d}{dy}\tan y=1+\tan^2y$, which can also be proved from first principles:$$\begin{align}\tan^\prime y&=\lim_{\Delta y\to0}\frac{\tan(y+\Delta y)-\tan y}{\Delta y}\\&=\lim_{\Delta y\to0}\frac{\frac{\tan y+\tan\Delta y}{1-\tan y\tan\Delta y}-\tan y}{\Delta y}\\&=\lim_{\Delta y\to0}\frac{(1+\tan^2y)\tan\Delta y}{\Delta y}\frac{1}{1-\tan y\tan\Delta y}\\&=(1+\tan^2y)\underbrace{\lim_{\Delta y\to0}\frac{\tan\Delta y}{\Delta y}}_1\underbrace{\lim_{\Delta y\to0}\frac{1}{1-\tan y\tan\Delta y}}_1,\end{align}$$where we use $\lim_{h\to0}\frac{\tan h}{h}=1,\,\lim_{h\to0}\tan h=0$.
A: $$
\lim_{\Delta x\to\ 0} \frac{\arctan(x+\Delta x)-\arctan x}{\Delta x} = \lim_{u\,\to\,v} \frac{u-v}{\tan u - \tan v} 
$$
Here we have $u = \arctan(x+\Delta x)$ so $\tan u = x+ \Delta x$ and so on.
Why is it that $u\to v$ as $\Delta x\to 0\text{?}$ In effect this says the arctangent function is continuous. That can only follow from properties of the tangent function on the interval $(-\pi/2,+\pi/2).$
Then
$$
\lim_{u\,\to\,v} \frac{u-v}{\tan u - \tan v} = \frac 1 {\sec^2 v} = \frac 1 {1+\tan^2 v} = \frac 1 {1+x^2.}
$$
Here of course you need to know how to differentiate the tangent function before doing this.
A: Switch to the straight function, find derivative and notice that
$$\dfrac{dy}{dx}=\dfrac{1}{dx/dy}=$$
So
$$x= \tan(y)$$
In the way you have obtained d.c. you can also find:
$$ \dfrac{dx}{dy} = \sec^2 y ={1+x^2}$$
$$ \dfrac{dx}{dy} = \dfrac{1}{\sec^2}  =\dfrac{1}{1+x^2}.$$
