Finding equation of tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$ 
Find the equation of the tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$

This is in the context of learning implicit differentiation.
First, I apply $\frac{dy}{dx}$ operator to both sides of the equation yielding:
$-\sin^{-2}(x) - \sin^{-1}(y)\frac{dy}{dx} = 0$
Second, I want to solve for $\frac{dy}{dx}$.
$\frac{dy}{dx} = -\sin^{-2}(x)\sin(y)$.
Third, I substitute the point $(0,\frac{1}{2})$ into the above equation to find the slope of the tangent line.
$\frac{dy}{dx}\mid_{(0,\frac{1}{2})} = -\sin^{-2}(0)\sin(\frac{1}{2}) = -0.479$
Finally, I substitute the slope into the point-slope equation of the line to obtain
$y = -0.479x + 0.2395$
Is this correct?
 A: Notice that $\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$. This can be seen because, by definition of an inverse function, $\sin(\arcsin(x)) =x$. So differentiating both sides gives
$$
1 = \frac{d}{dx}x =\frac{d}{dx}\sin(\arcsin(x)) \overset{\color{purple}{\text{Chain rule}}}{=} \cos(\arcsin(x))\frac{d}{dx}\arcsin(x) =\underbrace{\sqrt{1 - \color{blue}{\sin^2(\arcsin(x)}}}_{\sqrt{1 -\color{blue}{x^2}}}\frac{d}{dx}\arcsin(x)
$$
where on the last step we use the fact that $\sin^2(x) +\cos^2(x) = 1$ to write $\cos(x) = \sqrt{1 - \sin^2(x)}$.

With this established, implicitly differentiating $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ with respect to $x$ gives
$$
\frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} =0 \implies \frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}
$$
So substituting your point $\left(\color{purple}{x}, \color{blue}{y}\right) =\left(\color{purple}{0}, \color{blue}{\frac{1}{2}}\right)$ gives
$$
\frac{dy}{dx} = -\sqrt{\frac{1-\left( \color{blue}{\frac{1}{2}}\right)^2}{1-\left( \color{purple}{0}\right)^2}} = \color{green}{-\frac{\sqrt{3}}{2}}
$$
And by the slope-intercept form of a line, the desired tangent line is
$$
 \boxed{y = \color{green}{-\frac{\sqrt{3}}{2}}x  +   \color{blue}{\frac{1}{2}}}
$$
A: As
$$
\sin\left(\sin^{-1}x+\sin^{-1}y\right)=\sin\left(\frac{\pi}{6}\right)
$$
gives
$$
\sqrt{1-x^2} y+x \sqrt{1-y^2}=\frac{1}{2}
$$
calling $f(x,y) = \sqrt{1-x^2} y+x \sqrt{1-y^2}$ we have the tangent at $p_0=\left(0,\frac 12\right)$ as
$$
(p-p_0)\cdot\nabla f(p_0)=0
$$
with $p = (x,y),\ \ \nabla f(x,y) = \left\{\sqrt{1-y^2}-\frac{x y}{\sqrt{1-x^2}},\sqrt{1-x^2}-\frac{x y}{\sqrt{1-y^2}}\right\}$  giving
$$
\frac{\sqrt{3} x}{2}+y-\frac{1}{2}=0
$$
