# Differentiating \$ arctan (frac{a+x}{1-ax})- arctan x\$ Code Answer

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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.

\$\$arctan(dfrac{a+x}{1-ax}) = arctan{a}+arctan{x}\$\$ ðŸ˜‰

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 ðŸ™‚

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