If $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

Question

Aicharya · Accepted Answer

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} \sqrt{a^{\cos ^{-1} t}}}{\mathrm{dt}}}{\frac{\mathrm{d} \sqrt{a^{\sin ^{-1} t}}}{\mathrm{~d} t}}=\frac{\frac{1}{2 \sqrt{a^{\cos ^{-1} t}}} \cdot a^{\cos ^{-1} t} \cdot \log a \cdot \frac{-1}{\sqrt{1-t^2}}}{\frac{1}{2 \sqrt{a^{\sin ^{-1} t}}} \cdot a^{\sin ^{-1}} \cdot \log a \cdot \frac{1}{\sqrt{1-t^2}}}=\frac{-\sqrt{a^{\cos ^{-1} t}}}{\sqrt{a^{\sin ^{-1} t}}}=\frac{-y}{x}$

If $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

Solution