inverse trigonometric differentiation - 2 approaches, 2 different answers I have the following question:
$\arctan x^2+\arctan y^2=\frac{\pi}{4}$
Find $\frac{dy}{dx}$ in terms of $x$ and $y$.
I am aware we can differentiate implicitly straight away to get $\frac{2x}{1+x^4}+\frac{2y}{1+y^4}\cdot\frac{dy}{dx}=0$
then $\frac{dy}{dx}=\frac{-x\left(1+y^4\right)}{y\left(1+x^4\right)}$
However, I wasn't sure what was incorrect about the following approach and wondered if someone could highlight?
$\tan\left(\arctan x^2+\arctan y^2\right)=\tan\left(\frac{\pi}{4}\right)$
$\frac{x^2+y^2}{1-x^2y^2}=1$
$\frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(1-x^2y^2\right)$
$2x+2y\cdot\frac{dy}{dx}=-2xy^2-2x^2y\cdot\frac{dy}{dx}$
$\left(2+2x^2\right)y\cdot\frac{dy}{dx}=-2xy^2-2x$
$\frac{dy}{dx}=\frac{-2xy^2-2x}{\left(2+2x^2\right)\cdot y}=\frac{-x\left(1+y^2\right)}{y\left(1+x^2\right)}$

From $x^2 + y^2 = 1 - x^2y^2$, we get:
$$x^2 = \frac{1-y^2}{1+y^2} $$
Then,
$$\frac{1+y^2}{1+x^2} = \frac{1+y^2}{1+\frac{1-y^2}{1+y^2}} = \frac12(1+y^2)^2$$
And,
$$\frac{1+y^4}{1+x^4} = \frac{1+y^4}{1+ \left( \frac{1-y^2}{1+y^2} \right)^2} = \frac12 (1+y^2)^2$$
The two expressions are equal and there's nothing wrong with any of the two methods you used.
