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Given that $x$ and $y$ satisfy the equation:


Prove that, when $x=1, dy/dx=-1-\sqrt{3}/2$.

I tried to differentiate both sides:


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..

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Using the notation $y'=dy/dx$, the derivative of $\arctan(y)$ is $y'/(1+y^2)$ and not $1/(1+y^2)$. – Did Jul 7 '12 at 10:05
Since $y$ is a function of $x$, you're supposed to be using the chain rule if you're differentiating $\arctan\,y$... – J. M. Jul 7 '12 at 10:05
Your derivative for $\arctan\,y$ is still wrong. – J. M. Jul 7 '12 at 10:11
You should get $\displaystyle\frac{1}{1+x^2}+\frac{1}{1+y^2}\frac{dy}{dx}+\frac{y+x\frac{dy}{dx‌​}}{1+(xy)^2}=0$. – Paul Jul 7 '12 at 10:15
Vic. Please read my comment slow-ly. – Did Jul 7 '12 at 13:20
up vote 3 down vote accepted

All avatar noted; is completely the formal way for solving the problem. I wanted to show you another parallel approach. we know that $$\tan^{-1}(x)+\tan^{-1}(y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$ so try to convert your expression in terms of $x$ and $y$ first. This gives you an statement having $x$ and $y$ and some constants. Now, I think, differentiating in not hard to you. Notice that what avatar obtained above at first, is essentially here as well. :)

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Nicely done! +1 :^) – amWhy Mar 8 '13 at 1:43
@amWhy: Thank Amy. Ho do you do today? Was it good? – Babak S. Mar 8 '13 at 6:46

$x=1\implies \arctan 1+2\arctan y=11\pi/12=\arctan y=\pi/3\implies y=\sqrt 3$. Thus, taking derivative on both sides gives, $$ \begin{align} & \frac{1}{1+x^2}+\frac{1}{1+y^2}\frac{dy}{dx}+\frac{1}{1+x^2y^2}\left(y+x\frac{dy}{dx}\right)=0 \\[10pt] & \implies \frac{1}{2}+\frac{1}{4}\left(\frac{dy}{dx}\right)+\frac{1}{4}\left(\sqrt 3+\frac{dy}{dx}\right)=0 \\[10pt] & \implies 1/2\frac{dy}{dx}+(1/2+\sqrt 3/4)=0 \\[10pt] & \implies \frac{dy}{dx}=2(-1/2-\sqrt 3/4)=-1-\sqrt 3/2. \end{align} $$

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$$ \begin{align} & \arctan x + \arctan y + \arctan(xy) = \arctan\left( \frac{x+y}{1-xy} \right) + \arctan(xy) \\ & = \arctan\left( \frac{\frac{x+y}{1-xy} + xy}{1-\left(\frac{x+y}{1-xy}\right)xy} \right) = \arctan\left( \frac{ x + y + xy - x^2 y^2}{1-xy -x^2 y - xy^2} \right) = \frac{11}{12} \pi. \end{align} $$

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.

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