Given that $\tan^{-1}(x)+\tan^{-1}(y)+\tan^{-1}(xy)=11/12π$, prove that when $x=1, dy/dx=-1-\sqrt{3}/2$ Given that $x$ and $y$ satisfy the equation:
$$\arctan(x)+\arctan(y)+\arctan(xy)=11/12π$$
Prove that, when $x=1, dy/dx=-1-\sqrt{3}/2$.
I tried to differentiate both sides:
$$1/(1+x^2)+y/(1+y^2)+(y+x\,dy/dx)/(1+(xy)^2)=0$$
and I know that when $x=1, y=\sqrt{3}$ by putting $x=1$ into the given equation.
so I got $1/2+√3/4+(√3+dy/dx)/4=0$
$$\implies dy/dx=-2-2√3$$
Thanks for pointing out the mistake. but the answer is still wrong..

$$
\begin{align}
& \arctan x + \arctan y + \arctan(xy) = \arctan\left( \frac{x+y}{1-xy} \right) + \arctan(xy) \\
& = \arctan\left( \frac{\frac{x+y}{1-xy} + xy}{1-\left(\frac{x+y}{1-xy}\right)xy} \right) = \arctan\left( \frac{ x  + y + xy - x^2 y^2}{1-xy -x^2 y - xy^2} \right) = \frac{11}{12} \pi.
\end{align}
$$
So
$$
\frac{ x  + y + xy - x^2 y^2}{1-xy -x^2 y - xy^2} = \tan\left(\frac{11}{12}\pi\right) = -\tan\frac{\pi}{12} = \tan\left(\frac \pi 4 - \frac \pi 3\right)
$$
$$
x  + y + xy - x^2 y^2 = \left(1-xy -x^2 y - xy^2\right) \tan\left(\frac \pi 4 - \frac \pi 3\right)
$$
To differentiate both sides you need the chain rule and the product rule, but not the quotient rule.  And the tangent of that difference of two fractions is easy to find.
