MHT CET · Maths · Differentiation
If \(x=t+\frac{1}{t}\) and \(y=t-\frac{1}{t}\), the \(\frac{\mathrm{d} y}{\mathrm{~d} x}=\)
- A \(\frac{1-t^2}{1+t^2}\)
- B \(\frac{t^2+1}{t^2-1}\)
- C \(\frac{1+t^2}{1-t^2}\)
- D \(\frac{t^2-1}{t^2+1}\)
Answer & Solution
Correct Answer
(B) \(\frac{t^2+1}{t^2-1}\)
Step-by-step Solution
Detailed explanation
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} t}}{\frac{\mathrm{d} x}{\mathrm{~d} t}}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1}\)
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