Arctan addition equation (differential calculus) 
Example description:
Show with the help of differential calculus that:
$$\arctan x + \arctan\frac{1}{x}=\frac{\pi}{2} \qquad (x>0)$$

I know the rule of adding two tangents together
but if I add $x$ and $\dfrac 1x$ I get $\dfrac{x^2+1}x$.

How did my colleagues turn $\dfrac{\pi}2$ to $x^{-2}$ or $-\dfrac 1{x^2}$?
 A: The questionable assertions include two equivalences, and both of them appear to be completely false. The first one, as you point out, disappears the $\frac{\pi}2$ and somehow brings in $x^{-2}$. The second equivalence changes $x^{-2}$ to $-\frac 1{x^2}$, which introduces a stray minus sign. Whoever wrote that did not know what he was doing.
Here is an outline of a proper proof. Differentiate the expression $\arctan x+\arctan \frac 1x$ for positive $x$. You will need to use the chain rule on the second term, which explains part of the questionable line you showed us. You will get an answer of zero after you simplify, showing that the expression is a constant. Substitute some simple value of $x$ into the original expression to determine the constant, and you are done.
A: It may be better to mention the Mean Value theorem (MVT). As pointed above by the comments it is actually an application of the MVT. Applying the MVT to $f(x)=\arctan x+\arctan\frac{1}{x}$ for $x>0$ which is differentiable (and hence continous) on any finite closed interval in this domain, you get $f'(x)=\frac{1}{x^2+1}-\frac{1}{x^2+1}=0$ and so $f(x)=$Constant. Since $f$ is a constant function and $f(1)=\frac{\pi}{2}$, implies Constant $=\frac{\pi}{2}$. Thus, $f(x)=\frac{\pi}{2}$.
