Prove that if $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$ then $\frac{dy}{dx}$ = $-\frac{y}x$

Question

Prove:

If $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$ where $\sin^{-1}$ and $\cos^{-1}$ are inverse trig function, show that $\frac{dy}{dx}$ = $-\frac{y}x$

Unfortunately I don't see how this can be done. By differentiating x and y separately via parametric differentiation, I get the following two results:

$\frac{dx}{dt} = \frac{\log a}{2\sqrt{1 - t^2}}$

and

$\frac{dy}{dt} = -\frac{\log a}{2\sqrt{1-t^2}}$

Unfortunately I don't see how I can use that to arrive at $\frac{dy}{dx} = -\frac{y}x$

Answer 1

Note that the domain must be $0 \le t \le \dfrac{\pi}2$

Then $\displaystyle xy = \sqrt{a^{\arcsin t + \arccos t}} = a^{\pi / 4}$

So $x\;dy + y\; dx = 0$

$\dfrac{dy}{dx} = -\dfrac yx$

or

$x \dfrac{dy}{dx} + y = 0$

$\dfrac{dy}{dx} = -\dfrac yx$

Answer 2

HINT:

$$x^2=a^{\sin^{-1}t}\implies2\ln x=\ln a(\sin^{-1}t)$$

$$\implies2(\ln x+\ln y)=\ln a(\sin^{-1}t+\cos^{-1}t)$$

Now $\sin^{-1}t+\cos^{-1}t=\dfrac\pi2$