If $x = \sqrt{a^{\sin^{-1}t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$,show that $\frac{dy}{dx} = -\frac{y}{x}$.

Given equations are $x = \sqrt{a^{\sin^{-1}t}}$ and $y = \sqrt{a^{\cos^{-1}t}}$.
This can be written as $x = a^{\frac{1}{2}\sin^{-1}t}$ and $y = a^{\frac{1}{2}\cos^{-1}t}$.
Taking the natural logarithm on both sides for $x$:
$\ln x = \frac{1}{2}\sin^{-1}t \ln a$.
Differentiating with respect to $t$:
$\frac{1}{x} \frac{dx}{dt} = \frac{\ln a}{2} \cdot \frac{1}{\sqrt{1-t^2}} \implies \frac{dx}{dt} = \frac{x \ln a}{2\sqrt{1-t^2}}$.
Similarly,for $y$:
$\ln y = \frac{1}{2}\cos^{-1}t \ln a$.
Differentiating with respect to $t$:
$\frac{1}{y} \frac{dy}{dt} = \frac{\ln a}{2} \cdot \left(-\frac{1}{\sqrt{1-t^2}}\right) \implies \frac{dy}{dt} = -\frac{y \ln a}{2\sqrt{1-t^2}}$.
Now,using the chain rule $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$:
$\frac{dy}{dx} = \frac{-\frac{y \ln a}{2\sqrt{1-t^2}}}{\frac{x \ln a}{2\sqrt{1-t^2}}} = -\frac{y}{x}$.
Hence,$\frac{dy}{dx} = -\frac{y}{x}$ is proved.