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If $x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}$, show that $\frac{d \mathbf{y}}{\mathbf{d x}}=-\frac{\mathbf{y}}{x}$
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Let $x=\sqrt{s}, \quad s=a^u, u=\sin ^{-1} t$ $\frac{d x}{d s}=\frac{1}{2 \sqrt{s}}, \frac{d s}{d u}=a^u \log a, \frac{d u}{d t}=\frac{1}{\sqrt{1-t^2}}$ $\frac{d x}{d t}=\frac{d x}{d s} \times \frac{d s}{d u} \times \frac{d u}{d t}=\frac{\sqrt{a^{\sin ^{-1}}} \mathrm{t} \log a}{2 \sqrt{1-t^2}}$
Now, let $y=\sqrt{s}, s=a^u, u=\cos ^{-1} t$, $\frac{d y}{d s}=\frac{1}{2 \sqrt{s}}, \frac{d s}{d u}=a^u \log a, \frac{d u}{d t}=\frac{-1}{\sqrt{1-t^2}}$ $\frac{d y}{d t}=\frac{d y}{d s} \times \frac{d s}{d u} \times \frac{d u}{d t}=\frac{-\sqrt{a^{\cos ^{-1} t}} \log a}{2 \sqrt{1-t^2}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\sqrt{\mathrm{a}^{\cos ^{-1} \mathrm{t}}}}{\sqrt{\mathrm{a}^{\sin -1} \mathrm{t}}}=-\frac{\mathrm{y}}{\mathrm{x}}$
Now, let $y=\sqrt{s}, s=a^u, u=\cos ^{-1} t$, $\frac{d y}{d s}=\frac{1}{2 \sqrt{s}}, \frac{d s}{d u}=a^u \log a, \frac{d u}{d t}=\frac{-1}{\sqrt{1-t^2}}$ $\frac{d y}{d t}=\frac{d y}{d s} \times \frac{d s}{d u} \times \frac{d u}{d t}=\frac{-\sqrt{a^{\cos ^{-1} t}} \log a}{2 \sqrt{1-t^2}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{\sqrt{\mathrm{a}^{\cos ^{-1} \mathrm{t}}}}{\sqrt{\mathrm{a}^{\sin -1} \mathrm{t}}}=-\frac{\mathrm{y}}{\mathrm{x}}$
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