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


The given equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$

$x=\sqrt{a^{\sin ^{-1}},}$ and $y=\sqrt{a^{\cos ^{-1}},}$

$\Rightarrow x=\left(a^{\sin ^{-1} t}\right)^{\frac{1}{2}}$ and $y=\left(a^{\cos ^{-1} t}\right)^{\frac{1}{2}}$

$\Rightarrow x=a^{\frac{1}{2} \sin ^{-1} t}$ and $y=a^{\frac{1}{2} \cos ^{-1} t}$

Consider $x=a^{\frac{1}{2} \sin ^{-1} 1}$

Taking logarithm on both the sides, we obtain

$\log x=\frac{1}{2} \sin ^{-1} t \log a$

$\therefore \frac{1}{x} \cdot \frac{d x}{d t}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\sin ^{-1} t\right)$

$\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \cdot \frac{1}{\sqrt{1-t^{2}}}$

$\Rightarrow \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}}$

Then, consider $y=a^{\frac{1}{2} \cos ^{-1} t}$

Taking logarithm on both the sides, we obtain

$\log y=\frac{1}{2} \cos ^{-1} t \log a$

$\therefore \frac{1}{y} \cdot \frac{d y}{d x}=\frac{1}{2} \log a \cdot \frac{d}{d t}\left(\cos ^{-1} t\right)$

$\Rightarrow \frac{d y}{d t}=\frac{y \log a}{2} \cdot\left(\frac{-1}{\sqrt{1-t^{2}}}\right)$

$\Rightarrow \frac{d y}{d t}=\frac{-y \log a}{2 \sqrt{1-t^{2}}}$

$\therefore \frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{\left(\frac{-y \log a}{2 \sqrt{1-t^{2}}}\right)}{\left(\frac{x \log a}{2 \sqrt{1-t^{2}}}\right)}=-\frac{y}{x} .$

Hence, proved.

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