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

Toolbox:
• By chain rule we have $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx} Step 1: Let x=\sqrt s s=a^{\large u} u=\sin^{-1}t \large\frac{dx}{ds}=\frac{1}{2\sqrt s} \large\frac{ds}{du}$$=a^{\large u}$$\log a \large\frac{du}{dt}=\frac{1}{\sqrt {1-t^2}} Step 2: \large\frac{dx}{dt}=\frac{dx}{ds}$$\times$$\large\frac{ds}{du}\times$$ \large\frac{du}{dt}$
$\quad\;\;=\large\frac{1}{2\sqrt{a^{\Large u}}}$$\times a^{\large u}\log a\times \large\frac{1}{\sqrt{1-t^2}} \quad\;\;=\large\frac{1}{2\sqrt{a^{\Large \sin^{-1}t}}}$$\times a^{\large \sin^{-1}t}\log a\times \large\frac{1}{\sqrt{1-t^2}}$
$\quad\;\;=\large\frac{\sqrt{a^{\Large\sin^{-1}t}}\log a}{2\sqrt{1-t^2}}$
Step 3:
$y=\sqrt{\large a^{\cos^{-1}t}}$
$\large\frac{dy}{dt}=\frac{1}{2\sqrt{a\cos^{-1}t}}$$\times a^{\large\cos^{-1}t}\log a\times \big(\large\frac{-1}{\sqrt{1-t^2}}\big) \quad\;\;=-\large\frac{\sqrt{a^{\Large\cos^{-1}t}}\log a}{2\sqrt{1-t^2}} Step 4: \large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\quad\;\;=-\large\frac{\sqrt{a^{\Large\cos^{-1}t}}\log a}{2\sqrt{1-t^2}}\times \large\frac{2\sqrt{1-t^2}}{\sqrt{a^{\Large\sin^{-1}t}}\log a}$
$\quad\;\;=-\Large\frac{\sqrt{a^{\Large\cos^{-1}t}}}{\sqrt{a^{\Large\sin^{-1}t}}}$
$\quad\;\;=-\large\frac{y}{x}$