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Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$ Find $f_x(x,y)$

My attempt,

$$ \begin{aligned} f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\ &=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\ &=\frac{1+y^2}{1+y^2+x^2+x^2y^2} \end{aligned} $$

But the given answer is $\frac{1}{x^2+1}$.

How?

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Hint:

$$1+x^2+y^2+x^2y^2=(1+x^2)+y^2(1+x^2)=?$$

Alternatively, see Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

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  • $\begingroup$ Thanks ! I got it. $\endgroup$ – Mathxx Mar 27 '18 at 16:00
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Because $$\frac{1+y^2}{1+y^2+x^2+x^2y^2}$$

and

$$\frac{1}{x^2+1}$$

are the same thing.

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  • 1
    $\begingroup$ When $y \ne \pm i$ $\endgroup$ – Paul Sinclair Mar 27 '18 at 17:24
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Alternatively, denote: $x=\tan a; y=\tan b$. Then: $$z=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b}=a+b.$$ So: $$z_x=a_x=(\arctan x)'=\frac{1}{1+x^2}.$$

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