I have differentiated $$ arctan (frac{a+x}{1-ax})- arctan x$$ and got zero as the result. I don’t really understand why the derivative is zero, I think it’s because $ arctan (frac{a+x}{1-ax})- arctan x$ would give a constant as the result hence the first derivative becomes zero, but I cannot find a way to prove that this is the case.
Would anybody explain to me why the first derivative is zero?
Any help is appreciated, thanks in advance.
Answer
$$arctan(dfrac{a+x}{1-ax}) = arctan{a}+arctan{x}$$ 😉
if you don’t know why:
$$tan(a+b) = dfrac{tan a + tan b}{1-tan atan b} implies a + b = arctan left( dfrac{tan a+ tan b}{1-tan a tan b}right)$$
let’s $a =arctan u$ and $b = arctan v$ replace and done 🙂