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

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  • 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}$
$\quad\;\;=-\large\frac{\sqrt{a^{\Large\cos^{-1}t}}\log a}{2\sqrt{1-t^2}}\times \large\frac{2\sqrt{1-t^2}}{\sqrt{a^{\Large\sin^{-1}t}}\log a}$
$\quad\;\;=-\Large\frac{\sqrt{a^{\Large\cos^{-1}t}}}{\sqrt{a^{\Large\sin^{-1}t}}}$
$\quad\;\;=-\large\frac{y}{x}$
answered May 10, 2013 by sreemathi.v
 

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