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Question: Answered & Verified by Expert
If $x=t+\frac{1}{t}$ and $y=t-\frac{1}{t}$, the $\frac{\mathrm{d} y}{\mathrm{~d} x}=$
MathematicsDifferentiationMHT CETMHT CET 2022 (10 Aug Shift 2)
Options:
  • 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 $\frac{t^2-1}{t^2+1}$
Solution:
2622 Upvotes Verified Answer
The correct answer is: $\frac{t^2+1}{t^2-1}$
$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{~d} t}}{\frac{\mathrm{d} x}{\mathrm{~d} t}}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1}$

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