MHT CET · Maths · Differentiation
If \(x=\sqrt{a^{\sin ^{-1} t}}\) and \(y=\sqrt{a^{\cos ^{-1} t}}\), then \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)
- A \(\frac{y}{x}\)
- B \(\frac{-x}{y}\)
- C \(\frac{x}{y}\)
- D \(-\frac{y}{x}\)
Answer & Solution
Correct Answer
(D) \(-\frac{y}{x}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} \sqrt{a^{\cos ^{-1} t}}}{\mathrm{dt}}}{\frac{\mathrm{d} \sqrt{a^{\sin ^{-1} t}}}{\mathrm{~d} t}}=\frac{\frac{1}{2 \sqrt{a^{\cos ^{-1} t}}} \cdot a^{\cos ^{-1} t} \cdot \log a \cdot \frac{-1}{\sqrt{1-t^2}}}{\frac{1}{2 \sqrt{a^{\sin ^{-1} t}}} \cdot a^{\sin ^{-1}} \cdot \log a \cdot \frac{1}{\sqrt{1-t^2}}}=\frac{-\sqrt{a^{\cos ^{-1} t}}}{\sqrt{a^{\sin ^{-1} t}}}=\frac{-y}{x}\)
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