inverse trig functions derivative

f(x) = tan^-1(2x+1)      f(x)'  ?

$$\tan[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ ] = 2x+1 \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\
(1+\tan^2(\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ ) )
\times
\left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ \right]'
= 2
\\\\
\left[ 1+(2x+1)^2} \right]
\times
\left[\textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ \right]'
= 2 \\\\
\boxed{ \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(2x+1)} \ \right]'
= \frac{2} {1+(2x+1)^2} } }$$

.

Thanks for your answer. I really would love some explanation and if you can show me how to solve this problem by using u-substituiton I would appreciate.

Thanks again. :)

$$\\y=tan^{-1}(2x+1)\\\\
Let\;\; u=2x+1\qquad \frac{du}{dx}=2\\\\
y=tan^{-1}\;u\\\\
u=tan\;y\\\\
\frac{du}{dy}=sec^2y\\\\
\frac{dy}{du}=\frac{1}{sec^2y}\\\\$$

--------

$$\\y=tan^{-1}(2x+1)\\\\
Let\;u=2x+1\qquad \frac{du}{dx}=2\\\\
y=tan^{-1}u\\\\
u=tan\;y\\\\
\frac{du}{dy}=sec^2y\\\\
\frac{dy}{du}=\frac{1}{sec^2y}\\\\
\begin{array}{rll}
\frac{dy}{dx}&=&\frac{dy}{du}\times\frac{du}{dx}\\\\
&=&\frac{1}{sec^2y}\times 2\\\\
&=&\frac{2}{sec^2y}\\\\
&=&$refer to explanation below to help get to next step$\\\\
&=&\frac{2}{1+u^2}\\\\
&=&\frac{2}{1+(2x+1)^2}\\\\
&=&\frac{2}{4x^2+4x+2}\\\\
&=&\frac{1}{2x^2+2x+1}\\\\
\end{array}$$

--------------------

 tan y = u/1