If $x = (a^{\sin^{-1}t})^{1/2}$ and $y = (a^{\cos^{-1}t})^{1/2}$, show that $\frac{dy}{dx} = \frac{-y}x$ If $x = (a^{\sin^{-1}t})^{1/2}$ and $y = (a^{\cos^{-1}t})^{1/2}$, show that $\frac{dy}{dx} = -\frac{y}x$
I have tried solving this sum and led to nowhere, I am not able to eliminate a from this equation. 
Thanks in advance
 A: Let $x=a^{\frac12\arcsin(t)}$ and $y=a^{\frac12\arccos(t)}$.  

METHODOLOGY $1$:

Direct differentiation reveals
$$\frac{dx(t)}{dt}=x(t)\frac{\log(a)}{2\sqrt{1-t^2}}\tag1$$
and
$$\frac{dy(t)}{dt}=-y(t)\frac{\log(a)}{2\sqrt{1-t^2}}\tag 2$$
Then, using $(1)$ and $(2)$ and applying the chain rule yields
$$\bbox[5px,border:2px solid #C0A000]{\frac{dy}{dx}=\frac{dy}{dt}\left(\frac{dx}{dt}\right)^{-1}=-y/x}$$
as was to be shown!


METHODOLOGY $2$:

Alternatively, using the identity $\arcsin(t)+\arccos(t)=\pi/2$, we can write
$$x(t)y(t)=a^{\frac12\left(\arcsin(t)+\arccos(t)\right)}=a^{\pi/4}\tag 3$$
Differentiating $(3)$ reveals
$$x(t)\frac{dy(t)}{dt}+y(t)\frac{dx(t)}{dt}=0$$
whereupon rearranging we obtain
$$\bbox[5px,border:2px solid #C0A000]{\frac{\frac{dy(t)}{dt}}{\frac{dx(t)}{dt}}=\frac{dy}{dx}=-y/x}$$
as expected!
A: Notice that
$$
u:=\ln(xy)^2=\ln a^{\cos^{-1}t+\sin^{-1}t}=(\cos^{-1}t+\sin^{-1}t)\ln a
$$
and therefore
$$
du=\left(-\dfrac{1}{\sqrt{1-t^2}}+\dfrac{1}{\sqrt{1-t^2}}\right)\ln a\,dt=0.
$$
But
$$
du=2d\ln(xy)=2\dfrac{d(xy)}{xy}=2\dfrac{ydx+xdy}{xy}.
$$
It follows that
$$
xdy=-ydx,
$$
i.e.
$$
\dfrac{dy}{dx}=-\dfrac{y}{x}
$$
A: Hint:
$$2\ln(x)=\sin^{-1}t\cdot\ln a$$
$\sin^{-1}t+\cos^{-1}t=\frac\pi2$
