Impossible identity? $ \tan{\frac{x}{2}}$ $$\text{Let}\;\;t = \tan\left(\frac{x}{2}\right). \;\;\text{Show that}\;\dfrac{dx}{dt} = \dfrac{2}{1 + t^2}$$
I am saying that this is false because that identity is equal to $2\sec^2 x$ and that can't be equal. Also if I take the derivative of an integral I get the function so if I take the integral of a derivative I get the function also so the integral of that is $x + \sin x$ which evaluated at 0 is not equal to $\tan(x/2)$
 A: Note that you seemed to have attempted finding $$\dfrac{dt}{dx} = \dfrac{d}{dx}\left( \tan \left(\frac x2\right)\right)= \frac 12 \sec^2\left(\frac x2\right)$$ However, the task at hand is to differentiate $f(t)$ with respect to $t$: i.e.,  $$\bf \dfrac{dx}{dt}\neq \dfrac{dt}{dx}$$
To obtain $\bf \dfrac{dx}{dt}$, we need to express $x$ as a function of $t$:
$$\begin{align} t & = \tan\left(\frac x2\right) \\  \arctan t & = \arctan\left(\tan \left(\frac x2\right)\right) \\ \arctan t + n\pi & = \frac x2 \\ \\  2\arctan t + 2n\pi & = x\end{align}$$
NOW we can find $\dfrac{dx}{dt}$: $$\dfrac{dx}{dt} = \frac{d}{dt}(2\arctan t + \underbrace{2n\pi}_{\text{constant}}) = \dfrac{2}{1 + t^2}$$
A: Your confusion is that you're asked to find $\frac{dx}{dt}$, not $\frac{dt}{dx}$. Start with
$$
t=\tan\frac{x}{2}
$$
Differentiate both sides with respect to $t$. Using the chain rule twice you'll have
$$
\begin{align}
\frac{dt}{dt}=1&=\sec^2\frac{x}{2}\cdot\frac{d}{dt}\left(\frac{x}{2}\right)\\
&=\sec^2\frac{x}{2}\cdot\frac{1}{2}\cdot\frac{dx}{dt}
\end{align}
$$
so, multiplying both sides by 2 and dividing by $\sec^2(x/2)$ we get
$$
2\cos^2\frac{x}{2}=\frac{dx}{dt}
$$
and when $\tan\alpha= t$ we have $\cos\alpha = 1/\sqrt{t^2+1}$ (draw a triangle), so
$$
\cos\frac{x}{2}=\frac{1}{\sqrt{t^2+1}}
$$
so we have, from the above displayed equation,
$$
\frac{dx}{dt}=\frac{2}{t^2+1}
$$
as required.
This technique is known as implicit differentiation and is very useful in cases like this where you don't have a function defined in terms of the variable with which you want to differentiate.
A: Let us rewrite what you have with some added notation. You are given 
$$
t(x)=\tan(x/2),
$$
a function of $x$. You are asked to find the derivative of the function $x$ with respect to the variable $t$. You must first express $x$ as a function of $t$. As other have mentionned, taking $\tan^{-1}$ on both sides yields
$$
\frac{x}{2}=\tan^{-1}(t)+n\pi.
$$
Note that the $n\pi$ is there because of the periodicity of $\tan$, but won't change a thing when we differentiate with respect to $t$.
Now we have
$$
x(t)=\tan^{-1}(t)+2n\pi
$$
and we can find
$$
\frac{dx}{dt}=\frac{2}{1+t^2}
$$
provided you know the derivative of $\tan^{-1}(t)$.
A: You're wrong, the identity is correct.  Note that $t = \tan(x/2)$ implies
$x = 2 \arctan(t) + 2 n \pi$.
A: Paul. You've asked a couple of nice questions here, if I understand you right:

1) Given that $t$ can be expressed as a function of $x$ (specifically, you asked about the case where $t = \tan(\frac{x}{2})$), how can we find $\frac{dx}{dt}$, if it exists?
2) If we know that $\tan(\frac{x}{2}) = t$, then why does $x = 2\arctan(t) + 2n\pi$? 

The answer to the first question comes from the chain rule. By the chain rule, we can write
$$\frac{dx}{dx} = \frac{dx}{dt} \frac{dt}{dx};$$
since we know that $\frac{dx}{dx} = 1$, this tells us that
$$ 1 = \frac{dx}{dt} \frac{dt}{dx} \implies \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}}.$$
(This proof assumes that $x$ can be expressed as a differentiable function of $t$, but that assumption is valid here, so all is well.)
In your case, we see that 
$$ t = \tan(\frac{x}{2}) \implies \frac{dt}{dx} = \sec^2(\frac{x}{2}) \cdot \frac{1}{2} $$
by the chain rule, and so
$$\frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{2}{\sec^2(\frac{x}{2})}$$
But $\sec^2(x) = 1 + \tan^2(x)$, and so 
$$\frac{dx}{dt} = \frac{2}{\sec^2(\frac{x}{2})} = \frac{2}{1 + \tan^2(\frac{x}{2})} = \frac{2}{1 + t^2} $$
where to get the last equality I just plugged in $\tan(\frac{x}{2}) = t$.
Your second question has a fairly simple answer: the $\arctan(t)$ function as mathematicians usually define it outputs an angle $\theta$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, such that $\tan(\theta)$ = $t$. We have to bear in mind that, when we take $\arctan(\tan(\frac{x}{2}))$, we may not get back $\frac{x}{2}$ - just $\theta$. However, we know for certain that $\frac{x}{2} = \theta + n\pi$ for some integer $n$ and some $\theta$ between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ (ponder this if it is not clear). This $\theta$ will be precisely the one which $\arctan$ returns for the argument $t = \tan(\frac{x}{2})$, and so this $\theta$ is $\arctan(t)$. Thus,
$$\frac{x}{2} = \arctan(t) + n\pi$$
which is the second thing you were having trouble with.
Hope this helps! 
