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$\displaystyle \left[\arctan\left(\dfrac{x+a}{1-ax}\right)\right]'$



$\displaystyle \left[\arctan\left(\dfrac{x+a}{1-ax}\right)\right]'= \dfrac{1}{1+\left(\frac{x+a}{1-ax}\right)^2}\left[\dfrac{x+a}{1-ax}\right]'= \dfrac{(1-ax)^2}{(1-ax)^2+(x+a)^2}\dfrac{[x+a]'(1-ax)-(x+a)[1-ax]'}{(1-ax)^2}=\\ \qquad\displaystyle \dfrac{(1-ax)-(x+a)(-a)}{(1-ax)^2+(x+a)^2}= \dfrac{1+a^2}{(1-2ax+a^2x^2)+(x^2+2ax+a^2)}= \dfrac{1+a^2}{1+a^2x^2+x^2+a^2}= \dfrac{1+a^2}{(1+x^2)(1+a^2)}=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{\dfrac{1}{1+x^2}}$

How can this be possible? We know that $[\arctan x]'=\dfrac{1}{1+x^2}$... Are $\arctan x$ and $\arctan\left(\dfrac{x+a}{1-ax}\right)$ somehow related? Yes, it needs just a small trick. Remember that

$$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$

then, if $x=\tan\alpha$ and $a=\tan\beta$,

$$\arctan\left(\dfrac{x+a}{1-ax}\right)=\arctan\left(\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\right)=\arctan(\tan(\alpha+\beta))=\alpha+\beta=\arctan x+\arctan a$$

and $\arctan a$ is just a constant. Wow!