How to prove that $\frac{d\left(\tan^{-1}x\right)}{dx}=\frac{1}{1+x^2}$ directly using the definition. How to prove that $\frac{d\left(\tan^{-1}x\right)}{dx} =\frac{1}{1+x^2}$ using the definition.
I trying to prove some derivatives using fundamental theorems. But no idea of the proof $tan^{-1}x$
 A: We have $y=\tan^{-1}(x)$ if, and only if $\tan(y)=x$, $-\pi/2<x<\pi/2$. Applying implicit differentiation to the latter equation yields
\begin{align*}
\sec^2(y)\frac{dy}{dx}&=1\\
(1+\tan^2(y))\frac{dy}{dx}&=1\\
(1+x^2)\frac{dy}{dx}&=1\\
\frac{dy}{dx}&=\frac{1}{1+x^2}.
\end{align*}
A: $$\tan^{-1}(x+h)-\tan^{-1}x=\tan^{-1}\left(\dfrac{x+h-x}{1+(x+h)x}\right)$$
$$\implies\dfrac{\tan^{-1}(x+h)-\tan^{-1}x}h$$
$$=\dfrac{\tan^{-1}\left(\dfrac h{1+(x+h)x}\right)}{\dfrac h{1+(x+h)x}}\cdot\dfrac{{\dfrac h{1+(x+h)x}}}h$$
Set $\tan^{-1}\left(\dfrac h{1+(x+h)x}\right)=u$
As $h\to0\implies u\to0 $
A: For any bijective differentiable function $f:I\to J$, calling $g$ the reciprocal function, we have
$$\forall x\in I, g\left(f(x)\right)=x\tag{1}$$
(we have also $\forall x\in J, f\left(g(x)\right)=x.$)
From the definition, we remark that $g$ is differentiable and using the chain rule we get
$$\forall x\in I,\;g'(f(x))f'(x)=1.$$
Therefore, writing $y=f(x)\in J$ (such that $x=g(y)$), we obtain
$$g'(y)=\frac1{f'(x)}=\frac1{f'(g(y))}.$$
If $f=\tan$, $g=\arctan$, with $I=\mathbb R$ and $J=(-\frac\pi2,\frac\pi2)$,
it follows that
$$\arctan'(y)=\frac1{\tan'(\arctan y)}=\frac1{1+\tan^2(\arctan y)}=\frac1{1+y^2}.$$
A: Set $u=\arctan x, u+k=\arctan(x+h)$. This means $\tan u=x,\ \tan(u+k)=x+h$ and $-\dfrac\pi2<u, u+h<\dfrac\pi2$. Note that, by continuity, $h\to 0\iff k\to 0$.
Now the rate of variation can be rewritten as
$$\frac{\arctan (x+h)-\arctan x}{h}=\frac kh=\frac 1{\cfrac hk}=\frac 1{\cfrac{\tan(u+k)-\tan u}k},$$
so that 
\begin{align*}\lim_{h\to 0}&\frac{\arctan (x+h)-\arctan x}{h}=\lim_{k\to 0}\frac 1{\cfrac{\tan(u+k)-\tan u}k}\\={}&\frac 1{\lim\limits_{k\to 0}\cfrac{\tan(u+k)-\tan u}k}
=\frac1{\tan'u}=\frac1{1+\tan^2u}=\frac1{1+x^2}.
\end{align*}
A: Since several others have already provided a host of ways forward, I thought it might be instructive to present two more.  To that end, we proceed.

METHOD $1$: Using standard, non-calculus-based tools
In terms of the definition of the derivative, we have
$$\frac{d\arctan (x)}{dx}=\lim_{h\to 0}\frac{\arctan(x+h)-\arctan(x)}{h} \tag 1$$
Now, using the addition angle formula for the tangent function, $\tan(x-y)=\frac{\tan (x)-\tan(y)}{1+\tan(x)\tan(y)}$, we have
$$\frac{\arctan(x+h)-\arctan(x)}{h}=\frac{\arctan\left(\frac{h}{1+x(x+h)}\right)}{h} \tag 2$$
Now, we recall the basic inequality from geometry
$$|\sin x||\cos^2 x|\le |x\cos x|\le |\sin x| \tag 3$$
From $(3)$ it is easy to see that
$$\left|\frac{x}{1+x^2}\right|\le |\arctan (x)|\le |x| \tag 4$$
Using $(4)$ in $(2)$ reveals
$$\left|\frac{\frac{1}{1+x(x+h)}}{1+\left(\frac{h}{1+x(x+h)}\right)^2}\right| \le \left|\frac{\arctan\left(\frac{h}{1+x(x+h)}\right)}{h}\right|\le \left|\frac{1}{1+x(x+h)}\right|$$
whence applying the squeeze theorem yields the coveted result 
$$\lim_{h\to 0}\frac{\arctan(x+h)-\arctan(x)}{h}=\frac{1}{1+x^2}$$
And we are done!

METHOD $2$: Using the integral representation of the arctangent 
One way to define the arctangent is in terms of the integral
$$\arctan (x)=\int_0^x\frac{1}{1+t^2}\,dt$$
Then, we have
$$\frac{\arctan(x+h)-\arctan(x)}{h}=\frac1h \int_x^{x+h}\frac{1}{1+t^2}\,dt \tag 5$$
Applying the mean-value theorem to integrals to $(5)$, there is a number $x<\xi<x+h$ such that
$$\frac{\arctan(x+h)-\arctan(x)}{h}=\frac1h \frac{1}{1+\xi^2}((x+h)-x)=\frac{1}{1+\xi^2}$$
Then, using the squeeze theorem for $x<\xi<x+h$, we have
$$\lim_{h\to 0}\frac{\arctan(x+h)-\arctan(x)}{h}=\frac{1}{1+x^2}$$
as expected!
