Question

If x=(a^sin^-1 t)^1/2 ,y=(acos^-1 t)^1/2 show that dy/dx=-y/x

Answer 1

$x=\sqrt{a^{sin^{-1}t}}$
now differentiate x with respect to t,
$\frac{dx}{dt}=\frac{d(\sqrt{a^{sin^{-1}t}})}{dt}\\\\=\frac{1}{2\sqrt{a^{sin^{-1}t}}}.a^{sin^{-1}t}.loga.\frac{1}{\sqrt{1-t^2}}\\\\=\frac{\sqrt{a^{sin^{-1}t}}.loga}{2\sqrt{1-t^2}}\\\\=\frac{xloga}{2\sqrt{1-t^2}}----(1)$

similarly, $y=\sqrt{a^{cos^{-1}t}}$
now differentiate y with respect to t,
then we get,
$\frac{dy}{dt}=\frac{d(\sqrt{a^{cos^{-1}t}})}{dt}\\\\=\frac{1}{2\sqrt{a^{cos^{-1}t}}}.a^{cos^{-1}t}.loga.\frac{-1}{\sqrt{1-t^2}}\\\\=\frac{-yloga}{2\sqrt{1-t^2}}-----(2)$

now, dividing equations (2), by (1),
$\frac{dy}{dx}=\frac{\frac{-yloga}{2\sqrt{1-t^2}}}{\frac{xloga}{2\sqrt{1-t^2}}}\\\\=\frac{-y}{x}$

hence , dy/dx = -y/x [ proved/]