Given:
$[\tan^{-1}(x)]^2+[\cot^{-1}(y)]^2=1$
Find the tangent line equation to the graph at the point $(1,0)$ by implicit differentiation
I found the derivative:
$\dfrac{dy}{dx}=\dfrac{4\tan^{-1}(x)\cdot \cot^{-1}(y)}{(y^2+1)(x^2+1)}$
I may have done my derivative wrong, but my main concern is at some point $0$ will be plugged into $\cot$ inverse, resulting in division by zero.
I need help with this scenario.
$\displaystyle \frac{2\tan^{-1} x}{1+x^{2}} \, dx-\frac{2\cot^{-1} y}{1+y^{2}} \, dy=0$
$\displaystyle \frac{dy}{dx}=\frac{(1+y^{2})\tan^{-1} x}{(1+x^{2})\cot^{-1} y}$
Note that $(1,0)$ doesn't lie on the graph.
The graph can be parametrized as $(x,y)=(\tan \cos t,\cot \sin t)$